Key Concepts & Theorems at a Glance
- A circle is the set (locus) of all points in a plane at a fixed distance (the radius) from a fixed point (the centre).
- A chord is a line segment joining two points on a circle; a diameter is a chord through the centre — the longest possible chord.
- An arc is a connected portion of a circle between two points; the smaller piece is the minor arc, the larger is the major arc.
- Points on the same circle are concyclic. A quadrilateral whose vertices are concyclic is a cyclic quadrilateral.
Theorem 1: A unique circle passes through three non-collinear points (its circumcircle; the centre is the circumcentre).
Theorem 2: Equal chords subtend equal angles at the centre.
Theorem 3: Chords subtending equal angles at the centre are equal.
Theorem 4: The line from the centre to the midpoint of a chord is perpendicular to the chord.
Theorem 5: The perpendicular from the centre to a chord bisects the chord.
Theorem 6: Equal chords are equidistant from the centre.
Theorem 7: Chords equidistant from the centre are equal.
Theorem 8: The longer of two chords is closer to the centre.
Theorem 9: The angle subtended by an arc at the centre is double the angle it subtends at any point on the circle outside the arc. Corollary: the angle in a semicircle is 90°.
Theorem 10: If AB subtends equal angles at C and D on the same side of AB, then A, B, C, D are concyclic.
Theorem 11: Opposite angles of a cyclic quadrilateral sum to 180°.
Theorem 12 (converse): If opposite angles of a quadrilateral sum to 180°, its vertices are concyclic.
\[ \text{chord length } = 2\sqrt{r^2-d^2}, \quad \text{where } r=\text{radius},\ d=\text{perpendicular distance from centre} \]ActActivity: List some objects from nature that resemble a circle.
Many things in nature are circular or nearly circular in outline. Some examples:
The full moon and the sun;
ripples formed when a raindrop falls on still water;
the cross-section of a tree trunk (growth rings) or of a plant stem;
the centre/inflorescence of a sunflower;
a spider's orb web;
the pupil of an eye;
the cross-section of many fruits (e.g. an orange slice);
a rainbow (which is actually part of a circle);
the shape traced by the Earth's orbit around the Sun (approximately).
Think and Reflect
TRJamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?
Amina's suggestion uses the fact that a fold that makes the circular boundary overlap perfectly creates a crease along a diameter.
Fold the circular paper in half, so that the curved edge matches up with itself exactly on both sides. Open it out — the crease you see is a diameter of the circle.
Fold the paper again, in a different direction, so the boundary again overlaps perfectly, and open it out. This gives a second crease — another diameter.
Every diameter of a circle passes through the centre, so the point where the two creases (two different diameters) cross must be the centre of the circle.
Think and Reflect
TR1. What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?
2. What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?
3. The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points?
1. A square has rotational symmetry at 90°, 180°, 270° and 360° (4 rotational symmetries — it looks the same after each quarter-turn), and 4 lines of reflection symmetry (the two diagonals, and the two lines joining midpoints of opposite sides).
A regular pentagon has 5 rotational symmetries (at multiples of \(360^\circ\div5=72^\circ\): 72°, 144°, 216°, 288°, 360°), and 5 lines of reflection symmetry (each joining a vertex to the midpoint of the opposite side).
A regular hexagon has 6 rotational symmetries (at multiples of \(360^\circ\div6=60^\circ\)), and 6 lines of reflection symmetry (3 through pairs of opposite vertices, and 3 through midpoints of opposite sides).
In general, a regular n-gon has n rotational symmetries and n lines of reflection symmetry. A circle is the limiting case — it has symmetry for every angle of rotation, and every diameter is a line of symmetry.
2. The longest chord of a circle is its diameter, so for radius 5, the longest chord is \(2\times5=10\) units. There is no smallest chord — chords can be drawn as short as we like (by choosing the two endpoints closer and closer together), so their length can get arbitrarily close to 0, but a chord needs two distinct points, so there is no chord of length exactly 0, and no "smallest" chord that isn't beaten by an even shorter one.
3. The locus of points equidistant from two given points A and B is the perpendicular bisector of segment AB — a straight line, not a circle. Using the hint: let M be the midpoint of AB, and let P be any point on the perpendicular bisector. In triangles PMA and PMB: PM is common, AM = BM (M is the midpoint), and \(\angle PMA=\angle PMB=90^\circ\). By SAS congruence, \(\triangle PMA\cong\triangle PMB\), so \(PA=PB\). This shows every point on the perpendicular bisector is equidistant from A and B. Combined with the fact (given) that every point equidistant from A and B lies on the perpendicular bisector, the locus is exactly the perpendicular bisector of AB.
Think and Reflect
TR1. How many circles pass through two points on a plane?
2. Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?
3. As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?
4. As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?
5. You are given two points A and B on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?
Three of the infinitely many circles through fixed points A and B, centres sliding along the perpendicular bisector
1. Infinitely many circles pass through two given points A and B. Every point on the perpendicular bisector of AB can serve as the centre of one such circle (with radius equal to its distance to A, which equals its distance to B).
2. No — not every radius is possible. The smallest circle through A and B has AB itself as a diameter, giving the minimum radius \(\dfrac{AB}{2}\) (this happens when the centre is the midpoint of AB). There is no largest circle — as the centre moves further along the perpendicular bisector, the radius keeps growing without any upper limit, so the radius can be made as large as we like.
3. The radii increase — the farther the centre is from AB along the perpendicular bisector, the greater its distance to A (and B), so the radius grows.
4. The circle appears less curved. A bigger circle (larger radius) curves more gently near any given arc — think of how the Earth's surface looks almost flat locally because its radius is so large. So as the radius increases, the circle looks less curved (flatter) near A and B.
5. If A and B are simply required to lie somewhere on the boundary of a square (not necessarily at corners), infinitely many squares work, since A and B could sit anywhere along the sides of squares of many different sizes and orientations.
If A and B must be corners of the square, there are two cases. If AB is a side of the square, exactly 2 squares can be drawn — one on each side of line AB (the square "grows" either upward or downward from AB, using AB as one edge). If AB is a diagonal of the square, exactly 1 square can be drawn — the diagonal's midpoint is the square's centre, and the other diagonal (same length, perpendicular to AB through that midpoint) is completely determined, so there is only one such square.
Exercise Set 5.1
1Draw \(\triangle ABC\) with AB = 5 cm, \(\angle A = 70^\circ\) and \(\angle B = 60^\circ\). Draw the circumcircle of \(\triangle ABC\). Is the centre inside or outside the triangle?
Acute triangle ABC with circumcentre O inside the triangle
Since the angles of a triangle sum to 180°: \(\angle C = 180^\circ - 70^\circ - 60^\circ = 50^\circ\)
All three angles (70°, 60°, 50°) are less than 90°, so \(\triangle ABC\) is an acute-angled triangle.
To construct: draw AB = 5 cm. At A, construct a 70° angle; at B, construct a 60° angle. The rays from A and B meet at C, completing the triangle. Construct the perpendicular bisectors of any two sides (say AB and BC); their intersection is the circumcentre O. Draw the circle with centre O and radius OA.
2Draw \(\triangle ABC\) with AB = 5 cm, \(\angle A = 100^\circ\), AC = 4 cm. Draw the circumcircle of \(\triangle ABC\). Is the centre inside or outside the triangle?
Obtuse triangle ABC with circumcentre O outside the triangle
Here \(\angle A = 100^\circ\), which is greater than 90°, so \(\triangle ABC\) is an obtuse-angled triangle (obtuse at A).
To construct: draw AB = 5 cm. At A, construct a 100° angle, and mark point C on that ray so that AC = 4 cm. Join BC to complete the triangle. Construct the perpendicular bisectors of two sides; their intersection is the circumcentre O. Draw the circle with centre O and radius OA.
3Draw \(\triangle ABC\), with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of \(\triangle ABC\). Let the circumcentre be O. Measure OA, OB, OC.
Isosceles triangle ABC (BC = CA = 7 cm) with circumradii OA, OB, OC marked
This is an isosceles triangle (BC = CA = 7 cm). We can find the circumradius R using \(R=\dfrac{abc}{4\times\text{Area}}\), where a = 7, b = 7, c = 6 are the side lengths.
By Heron's formula, semi-perimeter \(s=\dfrac{6+7+7}{2}=10\)
Area \(=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{10(10-7)(10-7)(10-6)}=\sqrt{10\times3\times3\times4}=\sqrt{360}=6\sqrt{10}\approx18.97\text{ cm}^2\)
\(R=\dfrac{7\times7\times6}{4\times6\sqrt{10}}=\dfrac{294}{24\sqrt{10}}=\dfrac{49}{4\sqrt{10}}=\dfrac{49\sqrt{10}}{40}\approx3.87\text{ cm}\)
Since O is the circumcentre, OA, OB and OC are all radii of the same circumcircle, so they must all measure the same length when you actually measure them with a ruler after construction.
4What is the least possible radius of a circle through two points A and B?
As explored in the Think and Reflect above, the smallest circle through A and B is the one for which AB is a diameter — any smaller circle simply cannot reach both points.
Think, Draw and Infer
1A, B and C are three collinear points. Can you find a point P such that PA = PB = PC? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
Collinear points A, B, C on line ℓ — the perpendicular bisectors of AB and BC are both ⟂ ℓ, hence parallel, and never meet
No such point P exists. Here's why: since A, B, C lie on a single line \(\ell\), the perpendicular bisector of AB is a line perpendicular to \(\ell\) at the midpoint of AB, and the perpendicular bisector of BC is a line perpendicular to \(\ell\) at the (different) midpoint of BC.
Both perpendicular bisectors are perpendicular to the same line \(\ell\), so they are parallel to each other. Since A, B, C are distinct points in order, the midpoints of AB and BC are different points on \(\ell\), so these two parallel perpendicular bisectors are also distinct lines — and distinct parallel lines never meet.
Since a common centre P would have to lie on both perpendicular bisectors (being equidistant from A, B and from B, C), and these bisectors never intersect, no such point P exists.
Because there is no point equidistant from all of A, B, C, there is no circle that can have all three as its centre's equidistant points — so no circle passes through three collinear points.
This also answers the last part: since any 3 points that a line meets on a circle would have to be concyclic points lying on that line (collinear), and we've just shown collinear points can never lie on a common circle, a straight line can never cut a circle in three (or more) distinct points — a line meets a circle in at most 2 points.
2The circumcircle of a given \(\triangle ABC\) is drawn. Can there be other triangles congruent to \(\triangle ABC\) that share the same circumcircle?
△ABC (teal) rotated about the circumcentre O to give a congruent △A′B′C′ (dashed purple) on the same circumcircle
Yes. Rotating \(\triangle ABC\) about the circumcentre O by any angle produces a new triangle whose vertices still lie on the same circle (since rotation about the centre maps the circle to itself), and since rotation preserves distances, the new triangle is congruent to the original (same side lengths).
Similarly, reflecting \(\triangle ABC\) across any diameter of the circle also gives a congruent triangle inscribed in the same circumcircle.
ActExercise: A circle with centre O is drawn, and A, B, C, D are points on the circle (Fig. 5.19). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.
Fig. 5.19: circle with centre O; radii OA, OB (teal) and OC, OD (purple); K on arc AB, L on arc CD
This is a hands-on measuring activity — you draw the circle and points yourself, then use a protractor on your own figure, so the exact angle values depend on where you place A, B, C, D. Here is how to carry it out and interpret the result:
To find the angle subtended by arc AKB at the centre, join OA and OB, and measure \(\angle AOB\) — but measure it by sweeping along the same side as arc AKB (through K), not the short way round, in case the arc goes "the long way".
Compare the measured angle to 180°: if the swept angle is less than 180°, arc AKB is the minor arc between A and B; if it's more than 180° (a reflex angle), arc AKB is the major arc.
Do the same for arc CLD: measure \(\angle COD\), swept along the arc through L, and compare to 180°.
In a typical construction where K lies on the shorter path from A to B, arc AKB usually turns out to be the minor arc (central angle less than 180°); if L lies on the longer path from C to D, arc CLD usually turns out to be the major arc (central angle greater than 180°) — but this depends entirely on where you placed your points, so check your own measurements against the 180° rule.
Exercise Set 5.2
1Show that the triangle formed by a chord and the centre of the circle is isosceles.
Chord AB with centre C — radii CA, CB and chord AB form △CAB
Given: AB is a chord of a circle with centre C. Join CA and CB.
To prove: \(\triangle CAB\) is isosceles.
Proof: CA and CB are both radii of the same circle.
\(\therefore CA=CB\)
So \(\triangle CAB\) is isosceles, with AB as its base. Hence proved.
2Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Equal chords AB, DE (AB = DE) with centre C — △CAB (teal) and △CDE (purple)
Given: AB and DE are chords of the same circle with centre C, and \(AB=DE\).
To prove: \(\triangle CAB \cong \triangle CDE\).
Proof: In \(\triangle CAB\) and \(\triangle CDE\):
\(CA=CD\) (radii)
\(CB=CE\) (radii)
\(AB=DE\) (given)
\(\therefore \triangle CAB \cong \triangle CDE\) (SSS congruence). Hence proved.
Exercise Set 5.3
1Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that \(\angle CMA = \angle CMB = 90^\circ\). You need to show that AM = BM.)
Fig. 5.12: chord AB with midpoint M, centre C
Given: C is the centre; AB is a chord; CM ⟂ AB, so \(\angle CMA=\angle CMB=90^\circ\).
To prove: \(AM=BM\).
Proof: In \(\triangle CMA\) and \(\triangle CMB\):
\(CA=CB\) (radii)
\(CM=CM\) (common)
\(\angle CMA=\angle CMB=90^\circ\) (given)
\(\therefore \triangle CMA \cong \triangle CMB\) (RHS congruence)
\(\therefore AM=BM\). Hence proved. (This is Theorem 5.)
2An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.
Isosceles △ABC (AB = AC) inscribed in a circle with centre O — altitude AM (dashed) passes through O
Given: \(\triangle ABC\) is inscribed in a circle with centre O, and \(AB=AC\). M is the midpoint of BC.
To prove: The altitude from A to BC passes through O.
Proof: In \(\triangle ABM\) and \(\triangle ACM\):
\(AB=AC\) (given)
\(BM=CM\) (M is midpoint)
\(AM=AM\) (common)
\(\therefore \triangle ABM \cong \triangle ACM\) (SSS), so \(\angle AMB=\angle AMC\)
Also \(\angle AMB+\angle AMC=180^\circ\) (linear pair), so \(\angle AMB=\angle AMC=90^\circ\)
\(\therefore\) AM ⟂ BC at its midpoint M, i.e., AM is the perpendicular bisector of BC.
Also, \(OB=OC=r\) (radii), so O lies on the perpendicular bisector of BC.
Since AM is that same perpendicular bisector, O lies on AM. Hence proved.
3Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.
Two parallel chords on opposite sides of centre O — distance between midpoints = d₁ + d₂
For the 6 cm chord: half-length = 3 cm.
\(d_1=\sqrt{r^2-3^2}=\sqrt{25-9}=\sqrt{16}=4\text{ cm}\)
For the 8 cm chord: half-length = 4 cm.
\(d_2=\sqrt{r^2-4^2}=\sqrt{25-16}=\sqrt{9}=3\text{ cm}\)
Since the chords are on opposite sides of the centre:
Distance between midpoints \(=d_1+d_2=4+3=7\text{ cm}\)
Exercise Set 5.4
1Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.
Equal chords AB (teal) and FG (purple) — the perpendiculars from centre C to each are equal in length
Given: AB and FG are equal chords of a circle with centre C, radius r; \(AB=FG=2a\).
To prove: AB and FG are equidistant from C.
Proof: By Theorem 5, the perpendicular from C bisects each chord. For AB, in the right triangle formed by the radius, half-chord a, and distance d:
\(d^2+a^2=r^2 \Rightarrow d=\sqrt{r^2-a^2}\)
The same formula, with the same r and a, gives the distance for FG (since \(FG=2a\) too).
\(\therefore\) both chords are at distance \(\sqrt{r^2-a^2}\) from C. Hence proved.
2Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GF, and CE = CH, show that AB = GF.
Fig. 5.15: two chords AB, GF with perpendiculars CE, CH from centre C
Given: CE ⟂ AB, CH ⟂ GF, \(CE=CH\).
To prove: \(AB=GF\).
Proof: By Theorem 5, E is the midpoint of AB and H is the midpoint of GF.
In \(\triangle CEA\) and \(\triangle CHG\):
\(CA=CG\) (radii)
\(CE=CH\) (given)
\(\angle CEA=\angle CHG=90^\circ\)
\(\therefore \triangle CEA \cong \triangle CHG\) (RHS), so \(EA=HG\)
\(\therefore AB=2\,EA=2\,HG=GF\). Hence proved. (This is Theorem 7.)
3Solve the previous question using the Baudhāyana–Pythagoras theorem.
Same configuration as Fig. 5.15 — right triangles CEA and CHG give the Pythagorean route to AB = GF
In right \(\triangle CEA\): \(CA^2=CE^2+EA^2 \Rightarrow EA=\sqrt{r^2-CE^2}\)
In right \(\triangle CHG\): \(CG^2=CH^2+HG^2 \Rightarrow HG=\sqrt{r^2-CH^2}\)
Since \(CE=CH\): \(EA=HG\)
\(\therefore AB=2\,EA=2\,HG=GF\)
Exercise Set 5.5
1Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
Using chord length \(=2\sqrt{r^2-d^2}\) with \(r=7, d=6\):
Chord \(=2\sqrt{7^2-6^2}=2\sqrt{49-36}=2\sqrt{13}\text{ cm}\approx7.21\text{ cm}\)
2Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is \(2\sqrt{r^2-d^2}\).
Chord AB, centre C, foot of perpendicular M — right △CMA with CM = d, CA = r, MA = half the chord
Given: AB is a chord, C is the centre, CM ⟂ AB with \(CM=d\), and \(CA=r\).
Proof: By Theorem 5, M is the midpoint of AB. In right \(\triangle CMA\):
\(CA^2=CM^2+MA^2\)
\(r^2=d^2+MA^2\)
\(MA=\sqrt{r^2-d^2}\)
Since M is the midpoint, \(AB=2\,MA=2\sqrt{r^2-d^2}\). Hence proved.
3*In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.
No.
Counterexample: take \(r=5\). Let distance of CD \(=1\), so distance of AB \(=2\times1=2\).
\(CD=2\sqrt{25-1^2}=2\sqrt{24}\approx9.80\)
\(AB=2\sqrt{25-2^2}=2\sqrt{21}\approx9.17\)
Here \(2\,AB\approx18.33\), but \(CD\approx9.80\ne 2\,AB\).
This is because chord \(=2\sqrt{r^2-d^2}\) is not linear in d, so doubling d does not double or halve the chord length in a fixed ratio.
Exercise Set 5.6
1In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?
In \(\triangle OAB\), \(OA=OB=12\text{ cm}\) (both radii), and the angle between them, \(\angle AOB=60^\circ\).
Since \(\triangle OAB\) is isosceles with \(OA=OB\), the base angles are equal: \(\angle OAB=\angle OBA=\dfrac{180^\circ-60^\circ}{2}=60^\circ\)
So all three angles of \(\triangle OAB\) are 60° — the triangle is actually equilateral. This means \(AB=OA=OB=12\text{ cm}\)
2Let A and B be two points on a circle with centre O.
(i) Are there points X, Y on the circle, on the same side of AB, such that \(\angle AXB\) is different from \(\angle AYB\)?
(ii) Is it true that if \(\angle AXB = \angle AYB\), then X and Y lie on the same side of the circle?
(iii) If \(\angle AXB = \angle AYB\), and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?
Chord AB with X, Y both on the lower arc — same-side points always subtend equal angles
(i) No. By Theorem 9, the angle subtended by chord AB at any point on the same arc (same side of AB) always equals half the central angle of the arc on the other side. Since this value depends only on which arc the point lies on — not on the specific point — every point on the same side of AB subtends the exact same angle. So there are no such X, Y with different angles on the same side of the same circle.
(ii) Mostly yes, with one exception. If \(\angle AXB = \angle AYB\) and this common angle is not 90°, then X and Y must be on the same side of AB — because points on opposite arcs subtend supplementary angles (summing to 180°), and two supplementary angles can only be equal if each is 90°. So the one exception is when \(\angle AXB=\angle AYB=90^\circ\): this happens exactly when AB is a diameter, and in that case every point on the circle (on either side) subtends 90°, so X and Y could be on opposite sides too.
(iii) Yes, provided X and Y are on the same side of AB. This is exactly Theorem 10: if AB subtends equal angles at two points C, D on the same side of AB, then A, B, C, D are concyclic. So the circle through A, B, X automatically passes through Y too, as long as X and Y are on the same side of line AB.
3Find x in Fig. 5.26.
Fig. 5.26: Cyclic quadrilateral ADCB with ∠D = 100°, ∠B = x
ADCB is a cyclic quadrilateral (its four vertices lie on the circle), with D and B as a pair of opposite vertices.
By Theorem 11, opposite angles of a cyclic quadrilateral sum to 180°:
\(\angle D + \angle B = 180^\circ\)
\(100^\circ + x = 180^\circ\)
\(x = 80^\circ\)
ExExercise: A cyclic quadrilateral has angles measuring \(\angle A = 80^\circ\), \(\angle B = 110^\circ\), \(\angle C = 100^\circ\), and \(\angle D = 70^\circ\). Can such a quadrilateral be drawn? Explain why or why not.
By Theorem 12 (the converse of Theorem 11), a quadrilateral is cyclic exactly when both pairs of opposite angles sum to 180°.
Check opposite pair A, C: \(\angle A+\angle C=80^\circ+100^\circ=180^\circ\) ✓
Check opposite pair B, D: \(\angle B+\angle D=110^\circ+70^\circ=70^\circ+110^\circ=180^\circ\) ✓
Both conditions are satisfied. (As a sanity check, all four angles of any quadrilateral must sum to 360°, and indeed \(80+110+100+70=360^\circ\) ✓.)
End-of-Chapter Exercises
1In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
Chord \(=2\sqrt{r^2-d^2}=2\sqrt{13^2-5^2}=2\sqrt{169-25}=2\sqrt{144}=2\times12=24\text{ cm}\)
2An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
By Theorem 9, the angle subtended by an arc at the centre is double the angle it subtends at a point on the circle (outside the arc). So the angle at a point on the circle is half the central angle:
\(\dfrac{70^\circ}{2}=35^\circ\)
3The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.
Radius \(r=\dfrac{26}{2}=13\text{ cm}\). Half-chord \(=\dfrac{24}{2}=12\text{ cm}\)
By the Baudhāyana–Pythagoras theorem: \(d=\sqrt{r^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\text{ cm}\)
4A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?
Chord \(=2\sqrt{r^2-d^2}=2\sqrt{15^2-9^2}=2\sqrt{225-81}=2\sqrt{144}=2\times12=24\text{ cm}\)
5Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Chord AB with radii OA, OB (dashed) — the perpendicular bisector of AB (gold) passes through O
Given: AB is a chord of a circle with centre O.
To prove: O lies on the perpendicular bisector of AB.
Proof: \(OA=OB\) (radii), so O is equidistant from A and B.
The perpendicular bisector of AB is the locus of all points equidistant from A and B.
\(\therefore\) O lies on the perpendicular bisector of AB. Hence proved.
6The diameter of a circle is AB. Point C is on the circumference. What is the measure of the \(\angle ACB\)? Explain your reasoning.
AB is a diameter, C is on the circle — ∠ACB is always 90°
By the Corollary to Theorem 9, the angle subtended by a diameter at any point on the circle is 90°.
Reasoning: the arc from A to B (not containing C) subtends a straight angle (180°) at the centre, since AB is a diameter — moving from A to B along that arc sweeps out half the full circle. By Theorem 9, the angle at C (a point outside that arc) is half of this central angle: \(\dfrac{180^\circ}{2}=90^\circ\)
7ABCD is a cyclic quadrilateral inscribed in a circle. If \(\angle A\) measures 75°, what is the measure of \(\angle C\)? If \(\angle B\) measures 110°, what is the measure of \(\angle D\)?
By Theorem 11, opposite angles of a cyclic quadrilateral sum to 180°.
\(\angle A+\angle C=180^\circ \Rightarrow \angle C=180^\circ-75^\circ=105^\circ\)
\(\angle B+\angle D=180^\circ \Rightarrow \angle D=180^\circ-110^\circ=70^\circ\)
8Quadrilateral PQRS is inscribed in a circle. If \(\angle P = (2x+10)^\circ\) and \(\angle R = (3x-20)^\circ\), find the value of x and the measures of \(\angle P\) and \(\angle R\).
P and R are opposite vertices, so by Theorem 11: \(\angle P+\angle R=180^\circ\)
\((2x+10)+(3x-20)=180\)
\(5x-10=180\)
\(5x=190\)
\(x=38\)
\(\angle P=2(38)+10=76+10=86^\circ\)
\(\angle R=3(38)-20=114-20=94^\circ\)
Check: \(86^\circ+94^\circ=180^\circ\) ✓
9The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
Half-chord \(=\dfrac{16}{2}=8\text{ cm}\)
By the Baudhāyana–Pythagoras theorem: \(r=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{ cm}\)
10A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Kite ABCD: AB = DA = 5, BC = CD = 12, right angles at B and D, AC = 13 is a diameter
A cyclic quadrilateral with two pairs of adjacent equal sides (5, 5, 12, 12, in kite order) is a kite. A kite can only be cyclic if the two angles between its unequal sides are both 90° (this follows from Theorem 11: by the kite's symmetry, the two angles where a 5-side meets a 12-side are equal to each other, and since they're opposite angles of the cyclic quadrilateral, they must sum to 180° — so each is exactly 90°).
So the kite is made of two congruent right triangles, each with legs 5 and 12, joined along their common hypotenuse (which becomes a diagonal of the kite). By the Baudhāyana–Pythagoras theorem, that hypotenuse \(=\sqrt{5^2+12^2}=\sqrt{169}=13\) — and this diagonal is in fact a diameter of the circumcircle (since the angle in a semicircle is 90°, matching the right angles at those vertices).
Area of one right triangle \(=\dfrac{1}{2}\times5\times12=30\) square units
The kite is made of 2 such triangles, so total area \(=2\times30=60\) square units
11*Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?
Method: Construct the perpendicular bisectors of any two sides of the quadrilateral (say AB and CD) using ruler and compass. Their point of intersection is the circumcentre O (since O is equidistant from A, B and from C, D — same idea as Theorem 1).
Once O is located, check by inspection: if O lies within the boundary of the quadrilateral, the centre is inside; otherwise it is outside.
12*When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.
Equal chords AB, CD (AB = CD) meeting at P — the theorem shows AP = CP and PB = PD
Given: Chords AB and CD of a circle (centre O) intersect at P, with \(AB=CD\). M, N are midpoints of AB, CD.
To prove: \(AP=CP\) and \(PB=PD\).
Proof: Since \(AB=CD\), by Theorem 6: \(OM=ON\)
In \(\triangle OMP\) and \(\triangle ONP\):
\(OP=OP\) (common)
\(OM=ON\) (shown above)
\(\angle OMP=\angle ONP=90^\circ\) (Theorem 4)
\(\therefore \triangle OMP \cong \triangle ONP\) (RHS), so \(MP=NP\)
Also \(AM=\dfrac{AB}{2}=\dfrac{CD}{2}=CN\) (M, N are midpoints and \(AB=CD\))
\(\therefore AP=AM+MP=CN+NP=CP\), and similarly \(PB=PD\). Hence proved.
13*Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre. (Hint: Is it a circumcircle of a suitable triangle?)
Chord AB = 6 cm at perpendicular distance OM = 3 cm from centre O (radius = 3√2 cm)
Using chord \(=2\sqrt{r^2-d^2}\) with chord = 6, \(d=3\):
\(6=2\sqrt{r^2-9}\)
\(3=\sqrt{r^2-9}\)
\(r^2=18\)
\(r=3\sqrt2\text{ cm}\approx4.24\text{ cm}\)
Construction: Draw \(AB=6\text{ cm}\). Bisect AB at M and draw \(MO\perp AB\) with \(OM=3\text{ cm}\). With centre O and radius \(OA=\sqrt{OM^2+MA^2}=\sqrt{9+9}=3\sqrt2\text{ cm}\), draw the circle. AB is then a chord of length 6 cm at distance 3 cm from O.
14*Show that rectangle is the only parallelogram that can be inscribed in a circle.
Cyclic parallelogram ABCD — the proof forces every angle to 90°, so it must be a rectangle
Given: ABCD is a cyclic parallelogram.
To prove: ABCD is a rectangle.
Proof: Since ABCD is a parallelogram: \(\angle A=\angle C\), \(\angle B=\angle D\)
Since ABCD is cyclic, by Theorem 11: \(\angle A+\angle C=180^\circ\)
\(\therefore 2\angle A=180^\circ \Rightarrow \angle A=90^\circ\); similarly \(\angle B=90^\circ\)
All angles of ABCD equal 90°, so ABCD is a rectangle. Hence proved.
15*Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.
Rectangle ABCD inscribed in a circle — diagonals AC, BD (dashed) intersect at the centre O
Given: ABCD is a rectangle inscribed in a circle.
To prove: Diagonals AC, BD meet at the centre.
Proof: \(\angle ABC=90^\circ\) (angle of rectangle) is the angle subtended by AC at B.
\(\therefore\) AC is a diameter (converse of Corollary to Theorem 9).
Similarly, \(\angle BAD=90^\circ\) subtended by BD at A \(\Rightarrow\) BD is also a diameter.
Two diameters of a circle meet only at the centre.
\(\therefore\) AC and BD meet at the centre. Hence proved.
16*Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Three chords of equal length (gold), each with its midpoint (green dots) — all lying on a smaller concentric circle (dashed)
Let the circle have radius r, centre O; let all chords have fixed length L.
By Theorems 6–7, every chord of length L is at the same distance \(d=\sqrt{r^2-\left(\dfrac{L}{2}\right)^2}\) from O.
By Theorem 4, the midpoint of each chord is the foot of the perpendicular from O, so every midpoint is at distance d from O.
As the chord's direction varies, its midpoint sweeps out every point at distance d from O.
17*In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: "The centre of the circle lies on the angle bisector of \(\angle BAC\)".
Congruent chords AB = AC, with M, N the feet of the perpendiculars from O — △OMA ≅ △ONA
Given: \(AB=AC\), chords of a circle with centre O. M, N are midpoints of AB, AC.
To prove: AO bisects \(\angle BAC\).
Proof: Since \(AB=AC\), by Theorem 6: \(OM=ON\); also \(OM\perp AB\), \(ON\perp AC\) (Theorem 4).
In \(\triangle OMA\) and \(\triangle ONA\):
\(OA=OA\) (common)
\(OM=ON\) (shown above)
\(\angle OMA=\angle ONA=90^\circ\)
\(\therefore \triangle OMA \cong \triangle ONA\) (RHS), so \(\angle OAB=\angle OAC\)
\(\therefore\) AO bisects \(\angle BAC\), i.e., O lies on the angle bisector of \(\angle BAC\). Hence proved.
18Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.
Both chords on the same side of O — the 24 cm chord is closer, the 10 cm chord is farther, 7 cm apart
Half-chords: \(\dfrac{10}{2}=5\) and \(\dfrac{24}{2}=12\)
Let \(r\) be the radius. Distance from centre to the 10 cm chord: \(d_1=\sqrt{r^2-25}\). Distance from centre to the 24 cm chord: \(d_2=\sqrt{r^2-144}\)
Since the longer chord (24 cm) is closer to the centre (Theorem 8), and both chords are on the same side of the centre, the distance between them is the difference of the two distances: \(d_1-d_2=7\)
\(\sqrt{r^2-25}-\sqrt{r^2-144}=7\)
\(\sqrt{r^2-25}=7+\sqrt{r^2-144}\)
Squaring both sides: \(r^2-25=49+14\sqrt{r^2-144}+r^2-144\)
\(-25=49-144+14\sqrt{r^2-144}\)
\(-25=-95+14\sqrt{r^2-144}\)
\(70=14\sqrt{r^2-144}\)
\(5=\sqrt{r^2-144}\)
\(25=r^2-144\)
\(r^2=169\)
\(r=13\text{ cm}\)
Check: \(d_1=\sqrt{169-25}=\sqrt{144}=12\);
\(d_2=\sqrt{169-144}=\sqrt{25}=5\);
\(d_1-d_2=12-5=7\) ✓
19*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.
Regular hexagon inscribed in a circle — split into 6 equilateral triangles (one shaded)
Each side subtends a central angle of \(\dfrac{360^\circ}{6}=60^\circ\).
In \(\triangle OAB\) (O = centre, A, B adjacent vertices): \(OA=OB=r\), \(\angle AOB=60^\circ\), so base angles \(=\dfrac{180^\circ-60^\circ}{2}=60^\circ\) — the triangle is equilateral.
\(\therefore AB=r\)
Distance from centre to a side, using \(d=\sqrt{r^2-(\text{side}/2)^2}\):
\(d=\sqrt{r^2-\dfrac{r^2}{4}}=\dfrac{\sqrt3}{2}r\)
20A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about \(\angle MOP\) and \(\angle MNP\)? Explain your reasoning.
MNOP inscribed with MN a diameter; ∠MOP and ∠MNP both subtend chord MP (purple) from the same arc
\(\angle MOP\) and \(\angle MNP\) are both angles subtended by segment MP, at O and at N.
In the cyclic order M, N, O, P, chord MP has both N and O on the same arc.
\(\therefore \angle MOP=\angle MNP\) (angles in the same segment).
21Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., \(\angle CDE = \angle ABC\), where E is a point on the extension of side AD beyond D).
Cyclic quadrilateral ABCD with side AD extended to E — ∠CDE is the exterior angle at D
Since A, D, E are collinear (E lies on the extension of AD beyond D), \(\angle CDE\) and \(\angle CDA\) are angles on a straight line, so they are supplementary:
\(\angle CDE+\angle CDA=180^\circ\) …(1)
Since ABCD is a cyclic quadrilateral, by Theorem 11, opposite angles \(\angle CDA\) (i.e. \(\angle ADC\)) and \(\angle ABC\) sum to 180°:
\(\angle CDA+\angle ABC=180^\circ\) …(2)
From (1) and (2), both \(\angle CDE\) and \(\angle ABC\) equal \(180^\circ-\angle CDA\), so:
\(\angle CDE=\angle ABC\)
22*"There is no chord of a circle that is longer than its diameter." How do you justify this statement?
Chord \(=2\sqrt{r^2-d^2}\), where \(d\ge0\) is the distance from the centre.
This is maximised when \(d=0\) (its smallest possible value), giving chord \(=2\sqrt{r^2}=2r\), the diameter.
Since \(d\) cannot be negative, no chord can exceed \(2r\).
23*Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.
Point A inside the circle — the chord through A perpendicular to OA (teal) is shorter than any other chord through A (dashed purple)
Let \(OA=k\). Any chord through A has perpendicular distance \(d'\) from O, with \(d'\le OA=k\) (equality only when the chord ⟂ OA at A).
Chord \(=2\sqrt{r^2-d'^2}\), which decreases as \(d'\) increases.
So the chord is shortest when \(d'\) is largest, i.e. \(d'=k\), which happens exactly when the chord ⟂ OA at A.
24How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
Fig. 5.30: Angle in a semicircle, base angles a and b
Join OA (dashed). Since OA and the two base radii are all radii, both small triangles are isosceles, with base angles a and b as marked.
Angle at A \(=a+b\).
In the big triangle: \(a+b+(a+b)=180^\circ\)
\(2(a+b)=180^\circ\)
\(a+b=90^\circ\)
\(\therefore\) angle at A \(=90^\circ\).
25*In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C'D' is perpendicular to AB.
CC′ and DD′ ⟂ diameter AB; CD and C′D′ (dashed) have midpoints M, M′ — segment MM′ (red) is vertical, i.e. ⟂ AB
Given: \(CC'\perp AB\), \(DD'\perp AB\), where AB is a diameter. M, M′ are midpoints of CD, C′D′.
To prove: \(MM'\perp AB\).
Proof: Since AB passes through the centre and \(CC'\perp AB\), by Theorem 5, AB bisects \(CC'\) — so C, C′ are reflections of each other across AB.
Similarly D, D′ are reflections of each other across AB.
\(\therefore\) segment CD reflects to segment C′D′ across AB, so their midpoints M, M′ are also reflections of each other across AB.
A point and its reflection are always joined by a segment perpendicular to the mirror line.
\(\therefore MM'\perp AB\). Hence proved.
26*How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
Fig. 5.31: Cyclic quadrilateral ABCD with centre O joined to all vertices
Join O to A, B, C, D, forming 4 isosceles triangles (OA = OB = OC = OD = radii).
Let base angles be: \(\angle OAB=\angle OBA=p\);
\(\angle OBC=\angle OCB=q\);
\(\angle OCD=\angle ODC=u\);
\(\angle ODA=\angle OAD=v\)
\(\angle A=v+p,\ \angle B=p+q,\ \angle C=q+u,\ \angle D=u+v\)
Sum of angles of quadrilateral \(=360^\circ\):
\((v+p)+(p+q)+(q+u)+(u+v)=360^\circ\)
\(2(p+q+u+v)=360^\circ\)
\(p+q+u+v=180^\circ\)
\(\therefore \angle A+\angle C=(v+p)+(q+u)=180^\circ\), and \(\angle B+\angle D=(p+q)+(u+v)=180^\circ\)
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