Learning Objectives
After studying this chapter you should be able to:
- Define a circle precisely, and describe it as the locus of points at a fixed distance from a fixed point.
- Recognise the radius, chord and diameter of a circle, and use the complete rotational symmetry and the reflection symmetry of a circle.
- Explain why infinitely many circles pass through two points, and exactly one circle passes through three non-collinear points.
- Prove and apply theorems about equal chords, the perpendicular from the centre to a chord, and distances of chords from the centre.
- Compute chord lengths using the Baudhāyana–Pythagoras theorem: chord length \(2\sqrt{r^2-d^2}\).
- Distinguish minor and major arcs, and prove that the angle subtended by an arc at the centre is double the angle it subtends at any point on the remaining part of the circle.
- Apply the corollary that the angle subtended by a diameter at any point on the circle is \(90°\).
- Prove and apply results about concyclic points and cyclic quadrilaterals (opposite angles sum to \(180°\)).
1. The Circle and Its Parts
Circles appear all around us: when raindrops fall on water, in the cross-section of a plant stem, in the inflorescence of a sunflower, and in the full moon and the sun. Humans noticed long ago that every circle, big or small, has one common property: every circle has a centre, and all points on the circle are at equal distance from the centre. We turn this observation around and make it the definition of a circle.
Definitions
- Circle: the set of all points on the plane that are equidistant from a given point on that plane.
- Locus: the set of points that satisfy a given condition. A circle is thus the locus of points equidistant from a given point.
- Centre: the given point from which every point of the circle is equidistant.
- Radius: the distance from the centre to any point on the circle.
- Chord: a line segment joining two points of the circle.
- Diameter: a chord that passes through the centre of the circle.
In the figure, A is the centre. The angle subtended by the chord BC at the centre A is \(\angle BAC\). The chord DE passes through the centre, so it is a diameter. Notice that the diameter is the chord that contains the centre, and hence the chord that passes at zero distance from it.
Think and Reflect (finding the centre): Jamuna has a circular piece of paper and wants to find its centre. She folds the paper so that the two halves of the boundary overlap, presses the crease, and opens it. The crease is a line of reflection symmetry of the circle, so it passes through the centre. Folding again so that the boundary overlaps in a different way gives a second crease that also passes through the centre, so the two creases (diameters) meet exactly at the centre.
Solved Example 1
What is the length of the longest chord of a circle of radius 5 units? Is there a smallest chord?
Solution. The longest chord passes through the centre, so it is a diameter. Its length is twice the radius:
\[ \text{length of diameter} = 2 \times 5 = 10 \text{ units.}\]
As the two end points of a chord come closer and closer together, the chord becomes shorter and shorter; it can be made as close to 0 units as we wish, but it is never exactly 0 while the two end points are distinct. So there is no smallest chord with two distinct end points — chords of every positive length smaller than the diameter can be drawn.
Solved Example 2
Show that the triangle formed by a chord and the centre of the circle is isosceles.
Solution. Let BC be a chord of a circle with centre A. Then AB and AC are both radii of the circle, so \(AB = AC\). A triangle with two equal sides is isosceles, so \(\triangle ABC\) is isosceles, and BC is its base. The equal base angles are \(\angle ABC = \angle ACB\).
Solved Example 3
A circle has radius 3 cm. Where does each of the following points lie — on the circle, inside it, or outside it?
(a) A point P at distance 3 cm from the centre (b) A point Q at distance 2 cm (c) A point R at distance 4 cm
Solution. A point is on the circle exactly when its distance from the centre equals the radius.
- \(OP = 3\) cm = radius, so P lies on the circle.
- \(OQ = 2\) cm < radius, so Q lies inside the circle.
- \(OR = 4\) cm > radius, so R lies outside the circle.
2. Symmetries of a Circle
A rotating wheel looks the same at every moment: a point touching the ground now looks just like the point touching the ground later, and there is no way to tell them apart. We say the circle has complete rotational symmetry: rotate it by any angle about its centre and it looks exactly the same.
A circle also has reflection symmetry. Fold a circular paper so that the boundaries overlap; the crease is a line of reflection symmetry, and it passes through the centre. So the crease is a diameter. Every diameter is a line of reflection symmetry of the circle.
Think and Reflect — worked reasoning
(a) Rotational symmetries of regular polygons. A square looks the same under rotations of \(90°,\ 180°,\ 270°\) about its centre, so it has rotational symmetry of order 4, and it has 4 lines of reflection symmetry (two through opposite vertices and two through the midpoints of opposite sides). A regular pentagon has rotational symmetry of order 5 and 5 lines of reflection symmetry. A regular hexagon has rotational symmetry of order 6 and 6 lines of reflection symmetry.
(b) The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them: every point on the perpendicular bisector is equidistant from the two points, and every point that is equidistant from the two points lies on that perpendicular bisector.
3. How Many Circles Pass Through Two Points?
Given two points A and B, a circle passing through both has its centre O at equal distance from A and B. Since the perpendicular bisector of AB is exactly the locus of points equidistant from A and B, the centres of all circles through A and B lie on the perpendicular bisector of AB. There are infinitely many such points (and infinitely many such circles). If the centre is the midpoint of AB, the radius is \(AB/2\) and AB is a diameter; this is the smallest circle through A and B.
Solved Example 4
Two points A and B are 8 cm apart. What is the radius of the smallest circle passing through both A and B?
Solution. The centre of the smallest circle is the midpoint of AB, and AB is then a diameter. Hence
\[ \text{smallest radius} = \frac{AB}{2} = \frac{8}{2} = 4 \text{ cm.}\]
There is no largest circle through A and B: as the centre moves far away along the perpendicular bisector, the radius grows without bound and the circle becomes less and less curved.
Quick Practice
1. How many circles can be drawn through two given points? Infinitely many.
2. How many squares can be drawn with two given points A and B lying on the boundary? Infinitely many.
3. If A and B are to be adjacent corners of a square, how many squares can be drawn? Two — one on each side of AB.
4. A Unique Circle Through Three Non-Collinear Points
How many circles can be drawn through three points A, B and C? If the three points are collinear (on one straight line), no circle passes through all three — every line cuts a circle in at most two distinct points. For non-collinear points the answer is remarkable:
Theorem 1
There is a unique circle passing through three non-collinear points.
Why: Let the circle through A, B, C have centre O. Since \(OA = OB\), O lies on the perpendicular bisector of AB. Since \(OA = OC\), O lies on the perpendicular bisector of AC. The two perpendicular bisectors are not parallel (because A, B, C are non-collinear), so they meet at exactly one point — that point is O. Taking O as centre with radius \(OA\) gives the one and only circle through A, B and C.
Three non-collinear points determine a triangle, and the circle through its three vertices is special:
- The circle through the vertices A, B, C of \(\triangle ABC\) is the circumcircle; it is said to circumscribe the triangle, and the triangle is said to be inscribed in the circle.
- The centre O of the circumcircle is the circumcentre; it is the intersection of the perpendicular bisectors of the sides.
- For an acute-angled triangle the circumcentre lies inside the triangle (Fig. 5.5 style).
- For an obtuse-angled triangle the circumcentre lies outside the triangle.
- For a right-angled triangle the circumcentre is the midpoint of the hypotenuse.
Solved Example 5
A right-angled triangle has legs 6 cm and 8 cm. Find the radius of its circumcircle.
Solution. By the Baudhāyana–Pythagoras theorem, the hypotenuse is
\[ \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10 \text{ cm.}\]
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse, so the circumradius is half of it:
\[ r = \frac{10}{2} = 5 \text{ cm.}\]
Note: This is why the diameter of a circle is the hypotenuse — you always meet right angles at points on a circle (we will prove this in the corollary to Theorem 9).
Solved Example 6
Draw a triangle with \(AB = 5\) cm, \(\angle A = 100°\), \(AC = 4\) cm (Exercise Set 5.1, Q2). Without drawing, decide where its circumcentre lies.
Solution. Since \(\angle A = 100°\) is greater than \(90°\), the triangle is obtuse-angled (the other two angles must sum to \(80°\), so both are acute). The circumcentre of an obtuse-angled triangle lies outside the triangle.
5. Chords and the Angles They Subtend
Tie a thread at two points on a wheel and pull it tight; the thread is a chord of the wheel. Joining its end points to the centre, the thread subtends an angle at the centre. Now rotate the wheel: the thread rotates with it, so the new chord has the same length, and the angle it subtends at the centre is the same. The next two theorems make this precise.
Theorem 2
Equal chords of a circle subtend equal angles at the centre.
Why: Let AB and DE be equal chords. Then \(CA = CB = CD = CE = r\). Given \(AB = DE\), the triangles \(\triangle CAB\) and \(\triangle CDE\) are congruent by SSS congruence. Their corresponding angles are equal, so \(\angle ACB = \angle DCE\).
Theorem 3 (converse)
Chords that subtend equal angles at the centre are equal.
Why: Given \(\angle ACB = \angle DCE\) with all four end points on the circle, we have \(AC = DC\) and \(BC = EC\) (all radii). By SAS congruence, \(\triangle ACB \cong \triangle DCE\), so their corresponding sides are equal: \(AB = DE\).
Solved Example 7
A chord of a circle has length equal to the radius of the circle. Find the angle it subtends at the centre.
Solution. Join the ends of the chord to the centre. The three sides of the triangle so formed are equal (two radii and the chord), so the triangle is equilateral. Hence every angle is \(60°\), and the chord subtends
\[ \angle ACB = 60°.\]
Solved Example 8
Two chords of a circle subtend angles of \(40°\) and \(48°\) at the centre. Are the chords equal?
Solution. No. Equal chords subtend equal angles (Theorem 2); conversely, equal angles give equal chords (Theorem 3). Since the two angles are different, the chords are of different lengths — the chord subtending \(48°\) is longer.
6. Midpoints and Perpendicular Bisectors of Chords
What happens when we join the centre of a circle to the midpoint of a chord? The answer is a beautifully simple fact.
Theorem 4
The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
Why: Let M be the midpoint of chord AB of a circle with centre C. Since \(CA = CB\), the triangle \(\triangle CAB\) is isosceles, so its base angles are equal: \(\angle CAB = \angle CBA\).
We are given \(AM = BM\), and \(CA = CB\). By SAS congruence, \(\triangle CMA \cong \triangle CMB\). Hence \(\angle CMA = \angle CMB\).
But the two angles lie on a straight line, so they add to \(180°\). Two equal angles adding to \(180°\) must each be \(90°\). Therefore \(CM \perp AB\).
By asking the reverse question — what does the perpendicular from the centre to a chord do? — we get the converse (Exercise Set 5.3, Q1):
Theorem 5
The perpendicular from the centre of a circle to a chord bisects the chord.
Why: In Fig. 5.12-style figure, we are told \(\angle CMA = \angle CMB = 90°\). Also \(CA = CB\) (radii) and \(CM\) is common. The triangles \(\triangle CMA\) and \(\triangle CMB\) are congruent by RHS (right angle – hypotenuse – side), so \(AM = BM\). Hence M is the midpoint of AB.
Solved Example 9
Two parallel chords of lengths 6 cm and 8 cm lie on opposite sides of the centre of a circle of radius 5 cm. Find the distance between the midpoints of the chords.
Solution. The midpoint of each chord lies on the perpendicular from the centre to that chord (Theorems 4 and 5). Using the Baudhāyana–Pythagoras theorem on the right triangle formed by a radius, half the chord and the perpendicular:
For the 6 cm chord, half the chord is 3 cm, so
\[ d_1 = \sqrt{5^2-3^2} = \sqrt{25-9} = \sqrt{16} = 4 \text{ cm.}\]
For the 8 cm chord, half the chord is 4 cm, so
\[ d_2 = \sqrt{5^2-4^2} = \sqrt{25-16} = \sqrt{9} = 3 \text{ cm.}\]
The chords are on opposite sides of the centre, so the distance between their midpoints is \(d_1 + d_2 = 4 + 3 = 7\) cm.
7. Distance of Chords from the Centre
By the distance of a chord from the centre we always mean the perpendicular distance — the length of the perpendicular dropped from the centre to the chord (Theorem 5 guarantees this perpendicular lands on the midpoint).
A working formula
Let a circle of radius \(r\) have a chord whose perpendicular distance from the centre is \(d\), and let the chord length be \(c\). Theorem 5 cuts the chord into two equal halves of length \(c/2\), and together with the radius they form a right triangle:
\[ c = 2\sqrt{r^2-d^2}. \]
This is a direct application of the Baudhāyana–Pythagoras theorem: \(r^2 = d^2 + (c/2)^2\).
Solved Example 10
A chord of a circle is 5 cm away from the centre. If the radius is 13 cm, what is the length of the chord? (End-of-chapter exercise 1)
Solution. Using \(c = 2\sqrt{r^2-d^2}\) with \(r = 13\) and \(d = 5\):
\[ c = 2\sqrt{13^2-5^2} = 2\sqrt{169-25} = 2\sqrt{144} = 2 \times 12 = 24 \text{ cm.}\]
Solved Example 11
The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in it. Find the distance of the chord from the centre. (End-of-chapter exercise 3)
Solution. The radius is \(13\) cm, and half the chord is \(12\) cm. From \(r^2 = d^2 + (c/2)^2\):
\[ d = \sqrt{13^2-12^2} = \sqrt{169-144} = \sqrt{25} = 5 \text{ cm.}\]
Theorem 6
Chords of a circle having the same length are all at the same distance from the centre.
Why (first way): Let AB and FG be equal chords, and E, H their midpoints (Fig. 5.14 style). Triangles \(\triangle CAB\) and \(\triangle CFG\) are congruent by SSS (\(CA = CF\), \(CB = CG\), and \(AB = FG\)). The altitudes \(CE\) and \(CH\) of congruent triangles are equal, so \(CE = CH\).
Why (second way): After Theorem 5, \(AE = FH\) (halves of equal chords), \(\angle CEA = \angle CHF = 90°\), and \(CA = CF\). So \(\triangle CEA \cong \triangle CHF\) by RHS, giving \(CE = CH\).
Theorem 7 (converse)
Chords of a circle that are equidistant from the centre are equal in length. The two RHS-congruent triangles above can be run backwards: \(CE = CH\), \(CA = CF\) and right angles give \(AE = FH\), hence the chords are equal.
Theorem 8
Of two unequal chords, the longer chord is closer to the centre.
Why: Let CF and CG be the perpendiculars from the centre C to chords AB and DE, with AB longer, so \(AF > GD\) (F and G are the midpoints). Since AC and CD are radii, the Baudhāyana–Pythagoras theorem gives
\[ AC^2 = CF^2 + AF^2 \quad \text{and} \quad CD^2 = CG^2 + GD^2. \]
As \(AC = CD\), we get \(CF^2 + AF^2 = CG^2 + GD^2\). Now \(AF^2 > GD^2\), so it must be that \(CF^2 < CG^2\), i.e. \(CF < CG\). The longer chord AB is closer to the centre.
Solved Example 12
In a circle of radius 10 cm, two chords each lie at distance 6 cm from the centre. What is the length of each chord?
Solution. Since both chords have the same distance from the centre, by Theorem 7 they are equal. Their common length is
\[ c = 2\sqrt{10^2-6^2} = 2\sqrt{100-36} = 2\sqrt{64} = 16 \text{ cm.}\]
Solved Example 13
A circle has radius 15 cm and a chord is at distance 9 cm from the centre. What is the length of the chord? (End-of-chapter exercise 4)
Solution. \[ c = 2\sqrt{15^2-9^2} = 2\sqrt{225-81} = 2\sqrt{144} = 2 \times 12 = 24 \text{ cm.}\]
Solved Example 14
In a circle of radius 10 cm, chords of length 16 cm and 12 cm are drawn. Which chord is closer to the centre?
Solution. Their distances from the centre are: \[ d_{16} = \sqrt{10^2-8^2} = 6 \text{ cm}, \qquad d_{12} = \sqrt{10^2-6^2} = 8 \text{ cm.}\] Since \(6 < 8\), the 16 cm chord is closer to the centre — the longer chord is nearer the centre, exactly as Theorem 8 says.
Careful!
Equal angles at the centre imply equal chords (Theorem 3), but a chord at distance 0 from the centre is the diameter — the longest chord. If you push a chord away from the centre, its length shrinks; when its distance equals the radius, the chord collapses to a single point of length zero.
8. Arcs and the Angles They Make
An arc of a circle is a connected portion of the circle, defined by two end points and the curve along the circle between them. Two points A and B split the circle into two arcs: the smaller one is the minor arc and the bigger one is the major arc. The minor arc subtends the angle \(\angle AOB\) swept by turning from \(OA\) to \(OB\) along the minor arc; the major arc subtends the angle swept the other way (larger than \(180°\)).
Solved Example 15
An arc of a circle subtends an angle of \(70°\) at the centre. What is the measure of the angle subtended by this arc at a point on the circle? (End-of-chapter exercise 2)
Solution. Using the result of Theorem 9 (which follows), the angle at the centre is double the angle at the circle, so the angle at the point on the circle is
\[ \frac{70°}{2} = 35°.\]
9. The Angle Subtended by an Arc at a Point on the Circle
Take an arc AB and a point C on the circle, but outside the arc. Remarkably, \(\angle ACB\) does not depend on which point C we pick — any point on the circle and outside the arc gives the same angle. The key result is:
Theorem 9
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.
Why (first case): Let \(\angle ACB\) be the angle at the centre and D a point on the circle outside the arc, with DC extended to meet the circle again at E on the arc (Fig. 5.21 style). Since \(CB = CD\), \(\triangle DCB\) is isosceles and \(\angle CBD = \angle CDB\). The exterior angle \(\angle BCE\) of \(\triangle BCD\) satisfies
\[ \angle BCE = \angle CBD + \angle CDB = 2\angle BDC. \]
Similarly, \(CA = CD\) makes \(\triangle ACD\) isosceles and
\[ \angle ACE = \angle CAD + \angle CDA = 2\angle CDA. \]
Adding: \(\angle BCA = \angle BCE + \angle ECA = 2(\angle BDC + \angle CDA) = 2\angle BDA.\)
The second case (when the extension of DC meets the circle outside the arc) uses the same isosceles-triangle idea but subtracts the two small angles instead of adding, giving the same conclusion \(\angle BCA = 2\angle BDA\).
Corollary (a fact that follows from a proved result)
The angle subtended by a diameter at any point on the circle is \(90°\).
Why: Let AB be a diameter and D a point on the circle. The arc from A to B not containing D subtends the straight angle \(180°\) at the centre. By Theorem 9,
\[ \angle ADB = \frac{1}{2}\angle ACB = \frac{1}{2}\times 180° = 90°.\]
The equality of angles in the same segment is a beautiful distinguishing property of the circle (Fig. 5.23 style): for points D, E, F on the same arc AXB,
\[ \angle AEB = \angle ADB = \angle AFB = \frac{1}{2}\angle ACB. \]
If the points are inside or outside the circle, the angles differ (Fig. 5.25 of the book compares them); only points on the circle on the chosen arc give equal angles.
Solved Example 16
In a circle with centre O the central angle \(\angle AOB = 60°\), and the radius is 12 cm. What is the length of chord AB? (Exercise Set 5.6, Q1)
Solution. Join A and B to O. Since \(OA = OB = 12\) cm and \(\angle AOB = 60°\), the triangle is equilateral (isosceles with a \(60°\) vertex angle). Hence
\[ AB = OA = 12 \text{ cm.}\]
Solved Example 17
In Fig. 5.26, the four points A, B, C, D lie on a circle, \(\angle ADC = 100°\), and \(\angle ABC = x\). Find x.
Solution. The quadrilateral ABCD is cyclic, so its opposite angles are supplementary (Theorem 11):
\[ \angle ADC + \angle ABC = 180°\quad\Longrightarrow\quad 100° + x = 180°. \]
\[ x = 80°. \]
Alternative: the reflex angle that arc ABC subtends at the centre is \(2 \times 100° = 200°\), so the other central angle is \(160°\); half of it is \(x = 80°\).
Solved Example 18
The diameter of a circle is AB and C is a point on the circumference. What is the measure of \(\angle ACB\)? Explain. (End-of-chapter exercise 6)
Solution. AB is a diameter, so the arc not containing C subtends \(180°\) at the centre. By Theorem 9 (or its corollary),
\[ \angle ACB = \frac{1}{2} \times 180° = 90°.\]
So every angle subtended by a diameter is a right angle. The figure of exercise 24 shows this with two base angles a and b: in the two isosceles triangles \(a = a\) and \(b = b\), and \(2a + 2b = 180°\), giving \(a + b = 90°\).
10. Concyclic Points and Cyclic Quadrilaterals
Points that lie on the same circle are called concyclic. When the four vertices of a quadrilateral (a 4-gon) are concyclic, the quadrilateral is called a cyclic quadrilateral. Conditions for concyclicity are at the heart of this chapter.
Theorem 10
If a line segment AB subtends equal angles at two points C and D that lie on the same side of AB, then A, B, C, D are concyclic.
Why: Since A, B, C are non-collinear, Theorem 1 gives a unique circle through them. Suppose D is not on it — then D is either inside or outside. Join A to D and let the segment AD (or its extension) meet the circle at E.
If D is outside, \(\angle AEB\) is an exterior angle of \(\triangle BED\), so \(\angle AEB > \angle ADB\). But C and E lie on the same segment of the circle, so \(\angle AEB = \angle ACB\). This forces \(\angle ACB > \angle ADB\), contradicting the given \(\angle ACB = \angle ADB\). The same contradiction arises if D is inside. Hence D cannot be off the circle; it lies on the circle through A, B, C.
Theorem 11
The sum of two opposite angles of a cyclic quadrilateral is \(180°\).
Why: Let A, B, C, D be on a circle with centre O. Point A lies outside arc BCD, so \(\angle BAD\) is half the angle that arc BCD subtends at the centre — the reflex angle \(\angle BOD\). Similarly, \(\angle BCD\) is half the (non-reflex) angle \(\angle BOD\). Together they use up the complete angle at O:
\[ \angle BAD + \angle BCD = \frac{1}{2}\times 360° = 180°. \]
Theorem 12 (converse)
If two opposite angles of a quadrilateral add up to \(180°\), then the quadrilateral is cyclic.
Why (outline): Suppose ABCD has \(\angle BAD + \angle BCD = 180°\) but is not cyclic. Then A, D, B are non-collinear, so a circle passes through A, B, D; let it meet CD at E. Since ABED is cyclic, \(\angle BAD + \angle BED = 180°\) (Theorem 11). But then \(\angle BED = \angle BCD\), which is impossible because \(\angle BED\) is an exterior angle of \(\triangle BEC\) and hence strictly larger than \(\angle BCD\). So C must lie on the circle too.
Solved Example 19
ABCD is a cyclic quadrilateral. If \(\angle A = 75°\), what is \(\angle C\)? If \(\angle B = 110°\), what is \(\angle D\)? (End-of-chapter exercise 7)
Solution. Opposite angles of a cyclic quadrilateral are supplementary:
\[ \angle C = 180° - 75° = 105°, \qquad \angle D = 180° - 110° = 70°.\]
Solved Example 20
Quadrilateral PQRS is inscribed in a circle. If \(\angle P = (2x + 10)°\) and \(\angle R = (3x - 20)°\), find x and the measures of \(\angle P\) and \(\angle R\). (End-of-chapter exercise 8)
Solution. Opposite angles P and R of the cyclic quadrilateral sum to \(180°\):
\[ (2x + 10) + (3x - 20) = 180 \quad\Longrightarrow\quad 5x - 10 = 180 \quad\Longrightarrow\quad x = 38. \]
\[ \angle P = 2(38) + 10 = 86°, \qquad \angle R = 3(38) - 20 = 94°, \quad \text{and } 86° + 94° = 180°. \]
Solved Example 21
A quadrilateral has angles \(80°, 110°, 100°, 70°\). Can such a quadrilateral be inscribed in a circle?
Solution. Pair up opposite angles: \(80° + 100° = 180°\) and \(110° + 70° = 180°\). Since both pairs of opposite angles add to \(180°\), by Theorem 12 the quadrilateral is cyclic — yes, it can be drawn in a circle with these angles.
Solved Example 22
Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex equals the interior opposite angle, e.g. \(\angle CDE = \angle ABC\), where E is on the extension of side CD. (End-of-chapter exercise 21)
Solution. \(\angle CDE\) is an exterior angle, so \(\angle CDE = 180° - \angle BCD\). In the cyclic quadrilateral, opposite angles add to \(180°\), so \(\angle ABC + \angle BCD = 180°\), giving \(\angle ABC = 180° - \angle BCD\). Therefore \(\angle CDE = \angle ABC\).
Chapter Takeaway
Two winning ideas run through this whole chapter: (i) every chord plus two radii makes an isosceles triangle, so use the Baudhāyana–Pythagoras theorem freely; (ii) an angle at the centre is double an angle on the circle on the same arc — this one fact powers the semicircle corollary, the equal-angles-in-a-segment fact, and the cyclic quadrilateral theorems.
11. Applications of Circle Properties
Application 1 · Length of a walking path in a park
A circular park has radius 13 m. A straight walking path cuts across the park as a chord placed 5 m away from the centre. How long is the path?
Solution. \[ c = 2\sqrt{13^2-5^2} = 2\sqrt{144} = 2 \times 12 = 24 \text{ m.}\]
Application 2 · A chord on a circular board
A circular dartboard has radius 10 cm. A straight line of paint runs 8 cm from the centre. What is the length of the painted line?
Solution. \[ c = 2\sqrt{10^2-8^2} = 2\sqrt{36} = 2 \times 6 = 12 \text{ cm.}\]
Application 3 · Two parallel chords on a circular disc
Two parallel chords of lengths 6 cm and 8 cm are drawn on opposite sides of the centre of a circular disc of radius 5 cm. How far apart are the chords?
Solution. Half the lengths are 3 cm and 4 cm. Their distances from the centre are \(\sqrt{5^2-3^2} = 4\) cm and \(\sqrt{5^2-4^2} = 3\) cm. Being on opposite sides, the gap is \(4 + 3 = 7\) cm.
Application 4 · Decorative inlay on a round table
A circular table top has radius 25 cm. A decorative inlay is laid along a chord 7 cm from the centre. How long is the inlay?
Solution. \[ c = 2\sqrt{25^2-7^2} = 2\sqrt{625-49} = 2\sqrt{576} = 2 \times 24 = 48 \text{ cm.}\]
Application 5 · Circumcircle of a right-angled plaza
A triangular plaza has sides 60 m, 80 m and 100 m. A circular fountain ring is to be built so that it just passes through the three corners of the plaza. Find its radius.
Solution. Since \(60^2 + 80^2 = 3600 + 6400 = 10000 = 100^2\), the plaza is right-angled at the corner between 60 m and 80 m. Its circumcentre is the midpoint of the hypotenuse, so the required radius is half of 100 m: \(r = 50\) m. This is the radius of the ring through the three corners.
Application 6 · Angle of a decorative arc
A decorative arc on a circular clock face subtends \(50°\) at the centre. At what angle does this same arc appear at a point on the clock face?
Solution. Theorem 9: the angle at the centre is double the angle at the circle, so the required angle is \(50° \div 2 = 25°\).
Application 7 · Logo set inside a circle
A circular logo frames a cyclic quadrilateral. One pair of opposite angles of the quadrilateral measures \(4x°\) and \(80°\). Find x.
Solution. Opposite angles of a cyclic quadrilateral sum to \(180°\): \(4x + 80 = 180\), so \(4x = 100\), giving \(x = 25°\).
Application 8 · Measuring across a cylindrical duct
A cylindrical duct has a circular cross-section of radius 6 cm. A strip of tape, pulled tight across the section, lies 5 cm from the centre. What is the length of the strip under the tape (the chord)?
Solution. \[ c = 2\sqrt{6^2-5^2} = 2\sqrt{36-25} = 2\sqrt{11} \approx 6.6 \text{ cm.}\]
12. Competency-Based Questions
Competency 1
A circle of radius 13 cm has chords of lengths 14 cm and 24 cm. Without calculating both distances fully, which chord is closer to the centre? Verify your answer.
Answer. Because it is longer, the 24 cm chord is closer (Theorem 8). Checking: \(\sqrt{13^2-12^2} = 5\) cm for the 24 cm chord, and \(\sqrt{13^2-7^2} = \sqrt{120} \approx 10.95\) cm for the 14 cm chord. Since \(5 < 10.95\), the 24 cm chord is indeed closer.
Competency 2
The perpendicular from the centre O of a circle to a chord AB meets AB at M. Given \(AM = 6\) cm and \(OM = 8\) cm, find the radius of the circle.
Answer. By the Baudhāyana–Pythagoras theorem, \(OA = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\) cm.
Competency 3
Two chords of the same circle subtend angles of \(80°\) and \(90°\) at the centre. Which chord is longer? Give the reason.
Answer. The chord subtending \(90°\) is longer. Equal chords subtend equal angles; so if the angles differ, the chords differ — the larger central angle corresponds to the longer chord.
Competency 4
Points X and Y lie on the same arc of chord AB of a circle, and \(\angle AXB = 47°\). What is \(\angle AYB\)? Explain.
Answer. \(\angle AYB = 47°\). All angles in the same segment of a circle are equal (each equals half the angle subtended by arc AB at the centre).
Competency 5
ABCD is a cyclic quadrilateral and side AB is extended to E. If \(\angle ABC = 110°\), find \(\angle ADC\).
Answer. \(\angle ADC = 180° - 110° = 70°\) (opposite angles of a cyclic quadrilateral). Equivalently, the exterior angle at B equals the interior opposite angle at D.
Competency 6
A designer must make circular rings that pass through two fixed points 12 cm apart. On which set of points must the centres of all such rings lie? What is the smallest possible radius?
Answer. The centres lie on the perpendicular bisector of the 12 cm segment. The smallest ring uses the midpoint as its centre, giving the radius \(12/2 = 6\) cm; larger rings have centres farther along the perpendicular bisector.
13. More Solved Examples
Solved Example 23
Prove that the perpendicular bisector of a chord passes through the centre of the circle. (End-of-chapter exercise 5)
Proof. Let AB be a chord of a circle with centre C. Since \(CA = CB\), the centre C is equidistant from A and B. But the locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them, so C lies on the perpendicular bisector of AB. Hence the perpendicular bisector of any chord passes through the centre.
Solved Example 24
“There is no chord of a circle that is longer than its diameter.” Justify this statement. (End-of-chapter exercise 22)
Justification. Every chord not passing through the centre stands at some positive distance \(d > 0\) from the centre, so its length is
\[ 2\sqrt{r^2-d^2} < 2r. \]
The only chord at distance \(d = 0\) from the centre is the diameter, whose length is \(2r\). So the diameter is the longest possible chord.
Solved Example 25
A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle. (End-of-chapter exercise 19)
Solution. A regular hexagon is made of 6 congruent triangles from the centre, each with vertex angle \(360° \div 6 = 60°\) and two equal sides r. So each triangle is equilateral, and each side of the hexagon equals r.
The distance of a side from the centre is the altitude of the equilateral triangle: \[ \text{distance} = \sqrt{r^2 - \left(\frac{r}{2}\right)^2} = \sqrt{r^2 - \frac{r^2}{4}} = \frac{\sqrt{3}}{2}r. \]
Solved Example 26
A, B, C are three collinear points. Can you find a point P such that \(PA = PB = PC\)? What about the perpendicular bisectors of AB and BC? (Think, Draw and Infer)
Solution. If \(PA = PB\), then P lies on the perpendicular bisector of AB. If \(PB = PC\), then P also lies on the perpendicular bisector of BC. Since AB and BC lie on the same line, these two perpendicular bisectors are parallel, and parallel lines never meet. Hence no point P can be equidistant from all three collinear points — and, for the same reason, no circle passes through three collinear points (Theorem 1 always assumes non-collinearity).
14. Case-Based Questions
Case Study 1 · A circular garden
A circular garden of radius 13 m has a straight flower-bed chord placed 5 m from the centre, and a straight path chord placed 12 m from the centre.
(a) Find the length of the flower-bed chord.
(b) Find the length of the path chord.
(c) Which of the two chords is longer, and why?
Answers. (a) \(2\sqrt{13^2-5^2} = 2\sqrt{144} = 24\) m. (b) \(2\sqrt{13^2-12^2} = 2\sqrt{25} = 10\) m. (c) The flower-bed chord (24 m) is longer because a shorter distance from the centre gives a longer chord — the chord closer to the centre is the longer one (Theorem 8).
Case Study 2 · Two parallel rims of a wheel
Two parallel chords of lengths 10 cm and 24 cm are drawn on the same side of the centre of a wheel of radius r. The perpendicular distance between the two chords is 7 cm.
(a) Find the distances of the two chords from the centre in terms of r.
(b) Find the radius r of the wheel. (End-of-chapter exercise 18)
Answers. (a) Half lengths are 5 cm and 12 cm, so the distances are \(\sqrt{r^2-25}\) and \(\sqrt{r^2-144}\). (b) The longer (24 cm) chord is closer to the centre, so \(\sqrt{r^2-25} - \sqrt{r^2-144} = 7\). Trying r = 13: \(\sqrt{169-25}=12\) and \(\sqrt{169-144}=5\), and \(12-5=7\). So \(r = 13\) cm.
Case Study 3 · Cutting a circular plate
A circular plate of radius 10 cm is cut by two straight cuts (chords). One cut is at distance 6 cm from the centre and the other at distance 8 cm from the centre.
(a) Find the lengths of the two cuts.
(b) With the same radius, how should a cut be made to obtain the longest possible cut?
Answers. (a) At distance 6 cm: \(2\sqrt{10^2-6^2}=16\) cm; at distance 8 cm: \(2\sqrt{10^2-8^2}=12\) cm. (b) Pass the cut through the centre (distance 0), making it a diameter of length 20 cm.
Case Study 4 · A badge inside a circle
A school designs a badge in the shape of a cyclic quadrilateral with angles \(80°, x°, 100°, y°\) taken in order.
(a) Find x.
(b) Find y.
(c) State the property you used.
Answers. Opposite angles are supplementary: \(x = 180°-80° = 100°\) and \(y = 180°-100° = 80°\). (c) The sum of two opposite angles of a cyclic quadrilateral is \(180°\) (Theorem 11).
Case Study 5 · Ceiling dome with three exits
Three exit points on a circular hall mark a triangle. The building engineer wants the room’s circular chandelier ring to pass through all three exits.
(a) Which point of the triangle will be the centre of that ring?
(b) If the three exits form an obtuse-angled triangle, where will this centre lie?
(c) If the exits form a right-angled triangle with hypotenuse 14 m, compute the radius of the ring.
Answers. (a) The circumcentre — the intersection of the perpendicular bisectors of the sides (Theorem 1). (b) Outside the triangle (for an obtuse triangle). (c) The circumradius is half the hypotenuse: \(14/2 = 7\) m.
15. Common Mistakes and How to Avoid Them
| Mistake | Correct idea | Example / Fix |
|---|---|---|
| Calling any line in a circle a "diameter". | A diameter is a chord passing through the centre; it is the longest chord. | Radius 5 ⇒ diameter 10 units. |
| Thinking a circle passes through any three points. | Three collinear points have no circle through them. | A line meets a circle in at most two points. |
| Guessing where the circumcentre lies. | Acute ⇒ inside; obtuse ⇒ outside; right ⇒ midpoint of hypotenuse. | Sides 6, 8, 10 ⇒ circumradius 5, centre on hypotenuse. |
| Mixing up Theorems 4 and 5. | Centre-to-midpoint is perpendicular (T4); perpendicular from centre bisects (T5). | Both are used together in chord-distance problems. |
| Using the wrong formula sign. | Half-chord \(= \sqrt{r^2-d^2}\), so \(c = 2\sqrt{r^2-d^2}\). | r = 13, d = 5 ⇒ 24 cm (not \(\sqrt{25-169}\)). |
| Confusing "distance of a chord" with distance to an end point. | Distance means perpendicular distance from the centre to the chord. | Always drop the perpendicular (it hits the midpoint). |
| Assuming equal chords subtend equal angles anywhere. | Equal chords ⇔ equal angles at the centre. | Compare central angles, not angles on the circle. |
| Thinking the shorter chord is closer to the centre. | Of two unequal chords, the longer one is closer. | Diameter (longest) is at distance 0. |
| Forgetting there are two arcs. | Chords split the circle into major and minor arcs; angle at centre picks one. | Through points A and B, angles can be \(< 180°\) or \(> 180°\). |
| Applying "opposite angles ⇒ 180°" to any quadrilateral. | It holds only for cyclic quadrilaterals, and both pairs must be checked. | Check ∠A+∠C and ∠B+∠D. |
| Placing the 90° point on the diameter. | The point must be on the circle, not on the diameter. | Pick D on the semicircle, not between A and B. |
| Mishandling the factor 2 in Theorem 9. | Angle at centre = 2 × angle at circle; divide before you decide. | Central 70° ⇒ 35° at the circle. |
16. Exam Focus
- Know every theorem by heart and in words — Theorems 1 to 12 with their converses (2 ⇔ 3, 6 ⇔ 7) and the corollary on the angle in a semicircle.
- Practise writing "Given – To show – Why" for small proofs (isosceles triangle, exterior angle, SAS/SSS/RHS — that is the whole toolkit).
- Drill the chord formula \(c = 2\sqrt{r^2-d^2}\) and the reverse \(d = \sqrt{r^2-(c/2)^2}\); most numerical questions are one of these two.
- Keep the two arc sides in mind for Theorem 9: the angle at the circle uses the arc that is not containing the point.
- For cyclic quadrilaterals, always start with "opposite angles add to \(180°\)" and check both pairs when testing whether a quadrilateral is cyclic.
17. Quick Revision Cards
Circle
Set of all points in a plane equidistant from a fixed point (the centre). This is a locus.
Chord & diameter
Chord: segment joining two points of the circle. Diameter: chord through the centre; longest chord.
Rotational symmetry
A circle looks the same after rotation by any angle about its centre.
Reflection symmetry
Every diameter is a line of reflection symmetry (fold crease).
Two points
Infinitely many circles pass through two points; centres lie on the perpendicular bisector of the joining segment.
Three points
A unique circle passes through three non-collinear points — the circumcircle.
Circumcentre
Intersection of perpendicular bisectors. Acute: inside; obtuse: outside; right: midpoint of hypotenuse.
Equal chords
Equal chords subtend equal angles at the centre, and the converse (T2 ⇔ T3).
Midpoint line
Centre line to chord midpoint is perpendicular (T4); perpendicular from centre to chord bisects it (T5) — a chord+radius triangle is isosceles.
Equal distances
Equal chords are equidistant from the centre and conversely (T6 ⇔ T7).
Longer chord is closer
Of two unequal chords, the longer is nearer the centre (T8).
Chord formula
\(c = 2\sqrt{r^2-d^2}\), from the Baudhāyana–Pythagoras theorem.
Arcs
Two points split the circle into a minor arc and a major arc.
Angle at centre
Angle subtended by an arc at the centre is double the angle at any point on the remaining arc (T9).
Semicircle
Angle subtended by a diameter at any point on the circle is \(90°\).
Concyclic & cyclic
Equal angles on the same side of AB ⇒ concyclic (T10). Opposite angles of a cyclic quadrilateral sum to \(180°\) (T11 ⇔ T12).
18. Theorems at a Glance
| Theorem | Statement | How to use it |
|---|---|---|
| T1 | A unique circle passes through three non-collinear points. | Circumcircle & circumcentre; existence and uniqueness. |
| T2 | Equal chords subtend equal angles at the centre. | Compare central angles of equal chords (SSS proof). |
| T3 (converse) | Chords subtending equal angles at the centre are equal. | Equal central angles ⇒ equal chords (SAS proof). |
| T4 | The line from the centre to the chord midpoint is perpendicular to the chord. | Chord + radius triangle is isosceles; midpoint gives right angle. |
| T5 (converse) | The perpendicular from the centre to a chord bisects the chord. | Drop the perpendicular to bisect; paired with the chord formula. |
| T6 | Equal chords are equidistant from the centre. | Equal chords ⇒ equal perpendicular distances (SSS or RHS). |
| T7 (converse) | Chords equidistant from the centre are equal. | Equal distances ⇒ equal chords. |
| T8 | Of two unequal chords, the longer chord is closer to the centre. | Rank chords by distance; diameter is at distance 0. |
| T9 | Angle subtended by an arc at the centre is double the angle at any point on the remaining arc. | Central angle = 2 × angle on the circle; divide or multiply by 2. |
| Corollary | The angle subtended by a diameter at any point on the circle is \(90°\). | Any angle in a semicircle is a right angle. |
| T10 | If AB subtends equal angles at two points on the same side of AB, the four points are concyclic. | Prove four points lie on a circle (proof by contradiction). |
| T11 | Opposite angles of a cyclic quadrilateral sum to \(180°\). | Given one angle, find its opposite; build equations in x. |
| T12 (converse) | If two opposite angles of a quadrilateral sum to \(180°\), it is cyclic. | Test whether a quadrilateral can be inscribed in a circle. |
19. Key Definitions at a Glance
| Term | Meaning |
|---|---|
| Circle | Locus of points in a plane equidistant from a fixed point. |
| Centre | The fixed point from which every point of the circle is equidistant. |
| Radius | The distance from the centre to any point of the circle. |
| Chord | A line segment joining two points of the circle. |
| Diameter | A chord passing through the centre; its length is twice the radius. |
| Arc | A connected portion of the circle between two end points. |
| Minor / major arc | The smaller / larger of the two arcs between two end points. |
| Locus | The set of points satisfying a given condition. |
| Circumcircle | The circle through the vertices of a triangle; it circumscribes the triangle. |
| Circumcentre | The centre of the circumcircle; intersection of the perpendicular bisectors. |
| Inscribed | A triangle is inscribed in a circle when its vertices lie on the circle. |
| Concyclic points | Points that lie on the same circle. |
| Cyclic quadrilateral | A quadrilateral whose four vertices are concyclic. |
| Semicircle | The half of a circle cut by a diameter. |
20. How the Ideas Connect
The chapters’ main story: start with symmetries of the circle, use isosceles triangles (chord + two radii) and the Baudhāyana–Pythagoras theorem to understand chords, use arcs to relate angles at the centre with angles on the circle, and finish by deciding when points are concyclic — the property behind cyclic quadrilaterals.
21. Self-Practice Questions
SP 1
What is the length of the longest chord of a circle of radius 10 cm?
SP 2
A circle has radius 8 cm. Where do points at 8 cm, 6 cm and 10 cm from the centre lie?
SP 3
Two points A and B are 8 cm apart. How many circles pass through both? What is the radius of the smallest such circle?
SP 4
Can a circle pass through three distinct collinear points? Give a reason.
SP 5
State where the circumcentre of a triangle lies for an acute, an obtuse, and a right triangle.
SP 6
A right-angled triangle has legs 6 cm and 8 cm. Find the radius of its circumcircle.
SP 7
A chord of a circle has length equal to the radius. Find the angle it subtends at the centre.
SP 8
Two chords subtend \(70°\) and \(70°\) at the centre. Are the chords equal? Which theorem do you use?
SP 9
A chord lies 5 cm from the centre of a circle of radius 13 cm. Find its length.
SP 10
A chord of a circle of radius 13 cm has length 10 cm. Find its distance from the centre.
SP 11
A chord of length 6 cm is at distance 4 cm from the centre. Find the radius.
SP 12
Two parallel chords of lengths 8 cm and 6 cm are on the same side of the centre of a circle of radius 5 cm. Find the distance between the chords.
SP 13
A chord of length 48 cm is drawn in a circle of radius 25 cm. Find its distance from the centre.
SP 14
A circle of radius 15 cm has a chord at distance 9 cm from the centre. Find the chord length.
SP 15
A chord of length 30 cm lies in a circle of radius 17 cm. Find its distance from the centre.
SP 16
Two chords of a circle of radius 10 cm each lie at distance 6 cm from the centre. What length is each chord?
SP 17
In a circle of radius 10 cm, chords of lengths 16 cm and 12 cm are drawn. Which is closer to the centre?
SP 18
An arc subtends \(80°\) at the centre. What angle does it subtend at a point on the circle?
SP 19
An arc subtends \(25°\) at a point on the circle. What angle does it subtend at the centre?
SP 20
A circle has diameter 14 cm. What is the angle subtended by the diameter at any point on the circle?
SP 21
In a circle with centre O, \(\angle AOB = 60°\) and the radius is 6 cm. Find the chord AB.
SP 22
ABCD is cyclic with \(\angle A = 70°\) and \(\angle B = 88°\). Find \(\angle C\) and \(\angle D\).
SP 23
PQRS is cyclic with \(\angle P = (2x+40)°\) and \(\angle R = (x+20)°\). Find x, \(\angle P\) and \(\angle R\).
SP 24
A quadrilateral has angles \(40°, 140°, 90°, 90°\). Is it cyclic? Explain.
SP 25
State the exact conditions under which points A, B, C, D are concyclic (Theorem 10).
SP 26
Prove that the perpendicular bisector of a chord passes through the centre of the circle. (End-of-chapter exercise 5)
SP 27
Prove that the line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. (Theorem 4)
SP 28
Using the figure with centre O and base angles a and a, b and b, justify that the angle in a semicircle is \(90°\). (End-of-chapter exercise 24)
SP 29
Use a figure like Fig. 5.31 (angles p, q, u, v around the centre O) to justify that the sum of the opposite angles of a cyclic quadrilateral is \(180°\). (End-of-chapter exercise 26)
SP 30
A regular hexagon is inscribed in a circle of radius r. Find the length of its sides and the distance of each side from the centre. (End-of-chapter exercise 19)
SP 31
Let A be any point inside a circle with centre O. Show that the shortest chord through A is the one perpendicular to OA. (End-of-chapter exercise 23)
SP 32
Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre, 7 cm apart. Find the radius. (End-of-chapter exercise 18)
SP 33
The distance of a chord of length 16 cm from the centre is 6 cm. Find the radius. (End-of-chapter exercise 9)
SP 34
Use the Baudhāyana–Pythagoras theorem to show that equal chords of a circle are equidistant from the centre. (Exercise Set 5.4, Q1)
22. Self-Practice Answers
| No. | Answer |
|---|---|
| SP 1 | 20 cm (the diameter) |
| SP 2 | 8 cm → on the circle; 6 cm → inside; 10 cm → outside |
| SP 3 | Infinitely many; smallest radius 4 cm (AB as diameter) |
| SP 4 | No — a line meets a circle in at most two distinct points |
| SP 5 | Acute: inside; obtuse: outside; right: midpoint of hypotenuse |
| SP 6 | 5 cm (hypotenuse 10 cm, so circumradius 5 cm) |
| SP 7 | \(60°\) (triangle is equilateral) |
| SP 8 | Yes, equal (Theorem 2 / converse Theorem 3) |
| SP 9 | \(2\sqrt{13^2-5^2} = 24\) cm |
| SP 10 | \(\sqrt{13^2-5^2} = 12\) cm |
| SP 11 | \(\sqrt{3^2+4^2} = 5\) cm |
| SP 12 | 4 cm − 3 cm = 1 cm |
| SP 13 | \(\sqrt{25^2-24^2} = 7\) cm |
| SP 14 | \(2\sqrt{15^2-9^2} = 24\) cm |
| SP 15 | \(\sqrt{17^2-15^2} = 8\) cm |
| SP 16 | \(2\sqrt{10^2-6^2} = 16\) cm each (Theorem 7: they are equal) |
| SP 17 | The 16 cm chord (distance 6 cm < 8 cm) |
| SP 18 | \(40°\) |
| SP 19 | \(50°\) |
| SP 20 | \(90°\) |
| SP 21 | \(6\) cm (equilateral triangle) |
| SP 22 | \(\angle C = 110°\), \(\angle D = 92°\) |
| SP 23 | \(x = 40\); \(\angle P = 120°\), \(\angle R = 60°\) |
| SP 24 | Yes — \(40+140=180\) and \(90+90=180\) (Theorem 12) |
| SP 25 | C and D on the same side of AB, not on AB, with \(\angle ACB = \angle ADB\) |
| SP 26 | Use Theorem 5 & the idea of T1: the centre lies on both perpendicular bisectors AG of perpendicular bisectors; simply note the centre is equidistant from the chord’s end points, so it lies on the chord’s perpendicular bisector. |
| SP 27 | See Theorem 4 proof (isosceles triangle + SAS + angles on a line). |
| SP 28 | Two isosceles triangles give \(2a + 2b = 180°\), so \(a + b = 90°\). |
| SP 29 | Each interior angle is half the angle its opposite arc subtends at the centre; the two arcs cover \(360°\), so opposite angles sum to \(180°\). |
| SP 30 | Side length r; distance of each side \(\frac{\sqrt{3}}{2}r\) |
| SP 31 | Among chords through A the perpendicular distance d from O to the chord is largest when the chord is \(\perp OA\) (d = OA); chord length \(2\sqrt{r^2-d^2}\) is then smallest. |
| SP 32 | \(r = 13\) cm (since \(\sqrt{r^2-25}-\sqrt{r^2-144}=7\) gives r = 13) |
| SP 33 | \(\sqrt{8^2+6^2} = 10\) cm |
| SP 34 | Chord of length \(2a\) at distance d satisfies \(r^2 = a^2 + d^2\); equal chords have equal a, so equal d. |
Chapter Checklist
- I can define a circle precisely and explain the idea of a locus.
- I can identify radius, chord and diameter in a figure.
- I understand the complete rotational symmetry and the reflection symmetry of a circle.
- I can justify that infinitely many circles pass through two points, with centres on the perpendicular bisector.
- I can state and apply Theorem 1 (unique circle through three non-collinear points).
- I know where the circumcentre lies for acute, obtuse and right triangles.
- I can prove and apply Theorems 2 and 3 (equal chords ⇔ equal central angles).
- I can prove and apply Theorems 4 and 5 (perpendicular and bisector results for chords).
- I can compute chord lengths and distances using \(c = 2\sqrt{r^2-d^2}\).
- I can state Theorems 6, 7 and 8 and rank chords by their distance from the centre.
- I can distinguish minor and major arcs and use Theorem 9 (angle at centre is double).
- I can apply the corollary: angle in a semicircle is \(90°\).
- I can prove and apply Theorem 10 (concyclic points) and Theorems 11 and 12 (cyclic quadrilaterals).
- I always draw a figure and write "Given – To show – Why" for every proof.