Circles — Detailed Notes
5.1 Definitions
A circle is the set (locus) of all points in a plane that are at a fixed distance from a fixed point. The fixed point is the centre; the fixed distance is the radius. A chord is a segment joining two points on the circle; a chord through the centre is a diameter (the longest chord = 2 × radius). An arc is a connected part of the circle; the larger is the major arc, the smaller the minor arc.
5.2–5.3 Symmetry & circles through points
A circle has complete rotational symmetry about its centre and reflection symmetry about every diameter. Through two points A, B pass infinitely many circles — their centres all lie on the perpendicular bisector of AB. Through three non-collinear points there is exactly one circle (the circumcircle of the triangle); its centre, the circumcentre, is where the perpendicular bisectors of the sides meet. It lies inside an acute triangle, outside an obtuse triangle, and at the midpoint of the hypotenuse of a right triangle.
5.4–5.6 Chord theorems
• Equal chords subtend equal angles at the centre (and conversely). • The line from the centre to the midpoint of a chord is perpendicular to it; conversely, the perpendicular from the centre bisects the chord. • Equal chords are equidistant from the centre (and conversely). • Of two unequal chords, the longer chord is nearer the centre.
Chord length = 2√(r² − d²) (r = radius, d = distance of chord from centre)
5.7 Angles subtended by an arc
Theorem 9: the angle an arc subtends at the centre is double the angle it subtends at any point on the remaining circle. Consequences: angles in the same segment are equal; the angle in a semicircle is 90° (angle subtended by a diameter at any point of the circle).
5.8 Concyclic points & cyclic quadrilaterals
Points on one circle are concyclic. If a segment AB subtends equal angles at two points C, D on the same side of AB, then A, B, C, D are concyclic. A quadrilateral with all vertices on a circle is a cyclic quadrilateral, and in it the sum of each pair of opposite angles is 180° (and conversely, if opposite angles add to 180° the quadrilateral is cyclic). The exterior angle of a cyclic quadrilateral equals the interior opposite angle.
Key Terms
circle, centre, radius, chord, diameter, arc (major/minor), locus, perpendicular bisector, circumcircle, circumcentre, subtended angle, segment, concyclic, cyclic quadrilateral.
End-of-Chapter Exercises — Full Solutions
Complete solutions to all 26 end-of-chapter problems (starred * = HOTS), plus key in-text exercise answers. Many use the chord-length relation 2√(r²−d²).
1A chord is 5 cm from the centre; radius 13 cm. Find the chord length.
Half-chord = √(13² − 5²) = √(169 − 25) = √144 = 12. Chord = 2 × 12 = 24 cm.
2An arc subtends 70° at the centre. What angle does it subtend at a point on the circle?
The angle at any point on the remaining circle is half the central angle = 70° ÷ 2 = 35°.
3Diameter 26 cm, chord 24 cm. Find the distance of the chord from the centre.
Radius = 13 cm, half-chord = 12 cm. Distance = √(13² − 12²) = √(169 − 144) = √25 = 5 cm.
4Radius 15 cm; a chord is 9 cm from the centre. Find the chord length.
Half-chord = √(15² − 9²) = √(225 − 81) = √144 = 12. Chord = 24 cm.
5Prove that the perpendicular bisector of a chord passes through the centre.
Every point equidistant from the chord’s endpoints A and B lies on the perpendicular bisector of AB. The centre O satisfies OA = OB = radius, so O is equidistant from A and B — hence O lies on the perpendicular bisector of AB. Therefore the perpendicular bisector of any chord passes through the centre.
6AB is a diameter and C is on the circle. Find ∠ACB and explain.
∠ACB = 90°. The diameter AB subtends a central angle of 180°; by Theorem 9 the angle at a point on the circle is half of this, i.e. 90° (the angle in a semicircle).
7Cyclic quadrilateral ABCD: ∠A = 75°, ∠B = 110°. Find ∠C and ∠D.
Opposite angles of a cyclic quadrilateral add to 180°. ∠C = 180° − 75° = 105°; ∠D = 180° − 110° = 70°.
8PQRS cyclic with ∠P = (2x + 10)° and ∠R = (3x − 20)°. Find x, ∠P, ∠R.
∠P and ∠R are opposite, so (2x + 10) + (3x − 20) = 180 ⇒ 5x − 10 = 180 ⇒ 5x = 190 ⇒ x = 38. Then ∠P = 2(38) + 10 = 86° and ∠R = 3(38) − 20 = 94° (sum 180° ✓).
9A chord of length 16 cm is 6 cm from the centre. Find the radius.
Half-chord = 8 cm. Radius = √(8² + 6²) = √(64 + 36) = √100 = 10 cm.
10A cyclic quadrilateral has sides 5, 5, 12, 12. Find its area.
Take ABCD with AB = BC = 5 and CD = DA = 12 (a cyclic kite). By symmetry ∠A = ∠C, and being opposite they sum to 180°, so ∠A = ∠C = 90°. Then the diagonal BD = √(5² + 12²) = 13 (and is a diameter). Area = 2 × (½ × 5 × 12) = 60 square units.
11(*) For a cyclic quadrilateral, how can we tell whether the circumcentre lies inside or outside it?
Locate O as the meeting point of the perpendicular bisectors of the sides. O lies inside the quadrilateral when each side is a ‘minor’ chord — i.e. every arc between consecutive vertices is less than a semicircle. If one arc between two adjacent vertices exceeds 180° (equivalently, one interior angle is obtuse so its opposite side’s far arc is major), the centre falls outside. A quick check: if the quadrilateral has an obtuse angle, the centre lies outside on the far side of the side opposite that angle; if all four arcs are ≤ semicircles, O is inside.
12(*) Two equal chords intersect. Show the segments of one equal the corresponding segments of the other.
Let equal chords AB and CD meet at P. Equal chords are equidistant from the centre O; drop OM ⊥ AB and ON ⊥ CD, so OM = ON and M, N are the midpoints. In right triangles OMP and ONP, OP is common and OM = ON, so MP = NP. Since AM = CN (halves of equal chords), AP = AM ± MP and CP = CN ± NP give AP = CP and PB = PD. Thus the segments of one chord equal those of the other.
13(*) Draw a circle in which a chord of length 6 cm stands at distance 3 cm from the centre.
Radius = √(3² + 3²) = √(9 + 9) = √18 = 3√2 cm (≈ 4.24 cm). Construction: mark centre O; draw a circle of radius 3√2 cm; draw a line 3 cm from O — it meets the circle in two points 6 cm apart, giving the required chord (this is the circumcircle of the isosceles triangle formed by the chord and the two radii).
14(*) Show that a rectangle is the only parallelogram that can be inscribed in a circle.
In a parallelogram opposite angles are equal. In any cyclic quadrilateral opposite angles are supplementary. So each pair of opposite angles is both equal and sums to 180° ⇒ each angle = 90°. A parallelogram with all right angles is a rectangle. Hence only a rectangle can be inscribed.
15(*) Show that if a rectangle is inscribed in a circle, the intersection of its diagonals is the centre.
Each angle of the rectangle is 90°, so each diagonal subtends a right angle at the opposite vertex — meaning every diagonal is a diameter. The two diagonals, both diameters, pass through the centre, and since a rectangle’s diagonals bisect each other, their common midpoint is the centre of the circle.
16(*) Consider all chords of a circle of a fixed length. What shape do their midpoints form?
All equal chords are at the same distance d from the centre (where d = √(r² − (half-length)²)), and the midpoint of each chord is the foot of the perpendicular from O. So every midpoint is at distance d from O — the midpoints form a circle concentric with the original, of radius d.
17(*) Chords AB and AC are congruent. Explain why the centre lies on the bisector of ∠BAC.
In triangles OAB and OAC: OA common, OB = OC = radius, and AB = AC (given) ⇒ ΔOAB ≅ ΔOAC (SSS). Hence ∠OAB = ∠OAC, so AO bisects ∠BAC — the centre O lies on the angle bisector of ∠BAC.
18Two parallel chords 10 cm and 24 cm lie on the same side of the centre, 7 cm apart. Find the radius.
Half-chords 5 and 12; distances from centre √(r²−25) and √(r²−144) (the longer chord is nearer, so its distance is smaller). Same side ⇒ difference = 7: √(r²−25) − √(r²−144) = 7. Since (r²−25) − (r²−144) = 119 = 7 × (sum), the sum = 17; with difference 7 the two roots are 12 and 5. So r² − 25 = 144 ⇒ r² = 169 ⇒ r = 13 cm.
19(*) A regular hexagon is inscribed in a circle of radius r. Find its side and the distance of each side from the centre.
A regular hexagon splits into 6 equilateral triangles with the centre, so each side = r. The distance of a side from the centre (the apothem) = √(r² − (r/2)²) = √(3r²/4) = (√3/2)·r.
20Quadrilateral MNOP is inscribed in a circle with MN a diameter. What can you say about the angles at O and P?
A diameter subtends a right angle at any point on the circle. Since O and P lie on the circle, MN subtends 90° at each: ∠MON = 90° and ∠MPN = 90° (both are angles in a semicircle).
21Explain why the exterior angle of a cyclic quadrilateral equals the interior opposite angle.
At vertex D, the exterior angle ∠CDE and the interior angle ∠CDA form a linear pair: ∠CDE + ∠CDA = 180°. In the cyclic quadrilateral, ∠CDA + ∠ABC = 180° (opposite angles). Subtracting, ∠CDE = ∠ABC — the exterior angle equals the interior opposite angle.
22(*) Justify: no chord of a circle is longer than its diameter.
For a chord AB, consider triangle OAB with O the centre. By the triangle inequality AB ≤ OA + OB = r + r = 2r = diameter, with equality only when O lies on AB (i.e. AB is a diameter). So the diameter is the longest chord.
23(*) A is a point inside a circle with centre O. Show the shortest chord through A is perpendicular to OA.
Any chord through A is at some distance d from O, and its length = 2√(r² − d²) — shorter when d is larger. The distance d from O to a chord through A can be at most OA, and equals OA exactly when the foot of the perpendicular is A itself, i.e. when the chord is perpendicular to OA. That chord has the greatest d, hence the least length.
24Use the figure (angles a, b) to justify that the angle in a semicircle is 90°.
Join the point on the circle to the centre O. This makes two isosceles triangles (two sides are radii), with base angles a and b. The inscribed angle = a + b. The three angles of the big triangle give 2a + 2b = 180°, so a + b = 90°. Hence the angle in the semicircle = 90°.
25(*) Chords CC′ and DD′ are perpendicular to a diameter AB. Prove the segment joining the midpoints of CD and C′D′ is perpendicular to AB.
The diameter AB is an axis of symmetry of the circle. Since CC′ and DD′ are perpendicular to AB, reflection in AB maps C↔C′ and D↔D′. Therefore the midpoint of CD reflects to the midpoint of C′D′. A segment joining a point to its mirror image across AB is perpendicular to AB (and bisected by it). Hence MM′ ⊥ AB.
26Use the figure (angles p, q, u, v) to justify that opposite angles of a cyclic quadrilateral sum to 180°.
Each vertex angle is an inscribed angle equal to half the arc it ‘sees’. One pair of opposite angles together intercept the two arcs that make up the whole circle, so their sum = ½ × 360° = 180°. (In the figure, ∠A = ½(arc BCD) and ∠C = ½(arc BAD), and the two arcs complete the circle.)
IXKey in-text exercise answers (Sets 5.3, 5.5, 5.6).
Set 5.3 Q3 — chords 6 and 8, r = 5, opposite sides: distances √(25−9)=4 and √(25−16)=3, so distance between midpoints = 4 + 3 = 7 cm. Set 5.5 Q1 — r = 7, d = 6: chord = 2√(49−36) = 2√13 cm. Set 5.6 Q1 — central angle 60°, r = 12: the chord’s triangle with two radii and a 60° angle is equilateral, so the chord = 12 cm.