Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
| Result | Statement |
|---|---|
| Two points | Infinitely many circles; centres on the perpendicular bisector |
| Three points | Exactly one circle if non-collinear |
| Equal chords | Equal angles at centre; equal distances from centre |
| Chord length | l = 2√(r² − d²) |
| Longer chord | Nearer to the centre |
| Arc | Central angle = 2 × angle at circle |
| Semicircle | Angle = 90° |
| Cyclic quadrilateral | Opposite angles add up to 180° |
Try each problem first, then open the solution to check your working.
A circle has radius 10 cm. Find the length of a chord 6 cm from the centre.
Half chord = √(10² − 6²) = √64 = 8.
Chord = 16 cm.
Answer: 16 cm
A chord of length 24 cm is 5 cm from the centre. Find the radius.
Half chord = 12. r² = 12² + 5² = 169.
r = 13 cm.
Answer: 13 cm
In a circle of radius 10 cm there are parallel chords of lengths 16 cm and 12 cm. Find the distance between them if they are (i) on opposite sides of the centre (ii) on the same side.
Distances from the centre: 16-cm chord: √(100 − 64) = 6; 12-cm chord: √(100 − 36) = 8.
(i) Opposite sides: 6 + 8 = 14 cm.
(ii) Same side: 8 − 6 = 2 cm.
Answer: (i) 14 cm (ii) 2 cm
Chords of lengths 10 cm and 24 cm are drawn in the same circle. Which is nearer the centre?
The longer chord is nearer to the centre (Theorem 8).
So the 24 cm chord is nearer.
Answer: the 24 cm chord
A chord is as long as the radius r. Find its distance from the centre and the angle it subtends at the centre.
d = √(r² − (r/2)²) = (√3/2) r.
The triangle with two radii and the chord has all sides r (equilateral): the central angle is 60°.
Answer: d = (√3/2) r; 60°
A and B are 6 cm apart. How many circles of radius 5 cm pass through A and B? Where are their centres?
The centre lies on the perpendicular bisector of AB at distance 5 from A.
Distance from the midpoint M: √(5² − 3²) = 4. There are two such points, one on each side of AB.
Answer: two circles, centres 4 cm from AB on its perpendicular bisector
A right triangle has legs 6 cm and 8 cm. Find the radius of its circumcircle.
Hypotenuse = √(36 + 64) = 10.
The circumcentre is the midpoint of the hypotenuse, so R = 5 cm.
Answer: 5 cm
ΔABC is inscribed in a circle with centre O and ∠A = 50°. Find ∠BOC and ∠OBC.
∠BOC = 2 × 50° = 100° (Theorem 9).
ΔOBC is isosceles: ∠OBC = (180° − 100°)/2 = 40°.
Answer: ∠BOC = 100°; ∠OBC = 40°
An arc subtends 100° at the centre. Find the angle it subtends at a point on the remaining circle.
Angle at the circle = ½ × 100° = 50°.
Answer: 50°
ABCD is cyclic with ∠A = (3x + 10)° and ∠C = (5x − 30)°. Find x, ∠A and ∠C.
∠A + ∠C = 180°: 8x − 20 = 180, so x = 25.
∠A = 85°, ∠C = 95°.
Answer: x = 25; ∠A = 85°, ∠C = 95°
In a cyclic quadrilateral ABCD, ∠B = 110°. Find the exterior angle at D and ∠D.
∠D = 180° − 110° = 70°.
Exterior angle at D = 180° − 70° = 110° (equal to the interior opposite angle ∠B).
Answer: ∠D = 70°; exterior angle 110°
A quadrilateral has angles 70°, 100°, 110°, 80° in order. Can it be cyclic?
Opposite angles: 70° + 110° = 180° and 100° + 80° = 180°.
So by Theorem 12 it is cyclic.
Answer: yes
A parallelogram with sides 6 cm and 8 cm is inscribed in a circle. Find the radius.
Opposite angles of a parallelogram are equal and add up to 180°, so each is 90°: it is a rectangle.
Its diagonal √(36 + 64) = 10 is a diameter, so r = 5 cm.
Answer: 5 cm
AB subtends 40° at C on the circle. What does it subtend at D on the same side of AB, and at E on the opposite side?
Same segment: ∠ADB = 40°.
Opposite side: ∠AEB = 180° − 40° = 140° (A, E, B, C form a cyclic quadrilateral).
Answer: 40° and 140°
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Fold the circle exactly in half so the edges coincide; the crease is a diameter. Fold again along a different diameter. The two creases meet at the centre. (Equivalently: draw two chords and construct their perpendicular bisectors; they meet at the centre.)
Square: rotations by 90°, 180°, 270° (and 0°); 4 lines of reflection. Regular pentagon: rotations by multiples of 72°; 5 lines. Regular hexagon: rotations by multiples of 60°; 6 lines.
The longest chord is a diameter: 10 units. There is no smallest chord: chords can be made as short as we like, approaching zero length.
It is the perpendicular bisector of AB. Every point equidistant from A and B lies on it (given in the hint). Conversely, if P is on the perpendicular bisector with midpoint M, then ΔPMA ≅ ΔPMB (SAS: PM common, ∠PMA = ∠PMB = 90°, AM = BM), so PA = PB. So the bisector is exactly the locus.
Infinitely many: any point of the perpendicular bisector of AB is the centre of one.
The smallest radius is AB/2 (centre at the midpoint). Every radius larger than AB/2 occurs (twice: once on each side). There is no largest radius. No circle through A and B has radius less than AB/2.
They increase: r² = (AB/2)² + d², where d is the distance of the centre from AB.
Less curved: a larger radius gives a flatter circle.
Infinitely many squares have A and B on the boundary (the size can vary). With A and B as corners: if AB is a side there are 2 squares (one on each side of AB); if AB is a diagonal there is 1 square. So 3 squares in all.
The centre would have to lie on the perpendicular bisectors of AB and BC. For collinear points these bisectors are parallel (both perpendicular to the same line), so they never meet. Equivalently, a line meets a circle in at most two points.
Construction: draw AB = 5 cm; make 70° at A and 60° at B; the rays meet at C. Draw perpendicular bisectors of AB and AC; their intersection is O.
∠C = 180° − 70° − 60° = 50°, so the triangle is acute: the centre is inside. Radius R = AB/(2 sin 50°) ≈ 3.26 cm.
Construction: draw AB = 5 cm, 100° at A, mark C with AC = 4 cm; draw perpendicular bisectors of AB and AC.
∠A = 100° is obtuse, so the centre lies outside the triangle (beyond BC). BC ≈ 6.92 cm and R ≈ 3.52 cm.
Construct the triangle with compasses, then the perpendicular bisectors of AB and BC; they meet at O.
OA = OB = OC = the radius ≈ 3.87 cm (R = abc/4K with area K = √360 ≈ 18.97 cm²). The triangle is acute, so O is inside.
AB/2: the circle with AB as diameter. Any circle through A and B has radius at least half of AB (the chord AB cannot exceed the diameter).
No such P exists: P would lie on both perpendicular bisectors. They are both perpendicular to the same line, so they are parallel and distinct, and never meet. So no circle passes through three collinear points, and no line cuts a circle in three distinct points (at most two).
Yes. Rotate ΔABC about the circumcentre by any angle (or reflect it in a diameter): the image is congruent to ΔABC and its vertices stay on the same circle. There are infinitely many such triangles.
Let C be the centre and AB the chord. CA and CB are radii, so CA = CB. Hence ΔCAB is isosceles with base AB.
Both triangles have two sides equal to the radius r. If the bases are equal as well, then by SSS the triangles are congruent.
In Fig. 5.12, ∠CMA = ∠CMB = 90°, CA = CB (radii) and CM is common. By RHS, ΔCMA ≅ ΔCMB, so AM = BM.
In an isosceles triangle with AB = AC, the altitude from A is also the perpendicular bisector of BC. The centre O is equidistant from B and C (OB = OC = radius), so O lies on the perpendicular bisector of BC. So the altitude from A passes through O.
Distance of the 6 cm chord: √(25 − 9) = 4 cm. Distance of the 8 cm chord: √(25 − 16) = 3 cm.
The midpoints lie on the same line through the centre (perpendicular to both), on opposite sides: 4 + 3 = 7 cm.
Let a chord of length l be at distance d from the centre. The perpendicular bisects the chord (Theorem 5), so r² = d² + (l/2)², i.e. d = √(r² − l²/4). If two chords have the same length l, they have the same d. So equal chords are equidistant from the centre.
(The printed “GH” should be GF.) CE and CH are perpendiculars from C, so E and H are the midpoints of AB and GF. In right triangles CEA and CHF: CE = CH, CA = CF (radii), ∠CEA = ∠CHF = 90°. By RHS, AE = FH. Then AB = 2AE = 2FH = GF.
AE² = CA² − CE² and FH² = CF² − CH². Since CA = CF and CE = CH, AE² = FH², so AE = FH. As E, H are midpoints, AB = 2AE = 2FH = GF.
Half chord = √(7² − 6²) = √13. Chord = 2√13 cm ≈ 7.21 cm.
The perpendicular from the centre bisects the chord. In the right triangle formed by the radius to an end, the distance d and half the chord: r² = d² + (l/2)². So l/2 = √(r² − d²) and l = 2√(r² − d²).
Let CD be at distance d and AB at 2d. Then CD = 2√(r² − d²) and AB = 2√(r² − 4d²).
CD = 2AB would need r² − d² = 4(r² − 4d²), i.e. 15d² = 3r², d = r/√5. This holds only for that one value of d. For other values it fails (r = 5, d = 1: CD ≈ 9.80, AB ≈ 9.17).
So we cannot conclude CD = 2AB. We can only say CD is longer than AB.
Measure ∠AOB and ∠COD with a protractor, going from OA to OB via K and from OC to OD via L. If the angle is less than 180° the arc is a minor arc; if it is greater than 180° it is a major arc. (The values depend on your drawing.)
OA = OB = 12 and ∠AOB = 60°, so the base angles are 60° each: ΔOAB is equilateral. AB = 12 cm.
No. Both equal half the central angle of arc AB (Theorem 9), so they are equal.
Points on opposite sides give angles that add up to 180°. So equal angles on opposite sides force each to be 90°, which happens only when AB is a diameter. So yes, except when AB is a diameter (then X and Y can be on opposite semicircles, both with 90°).
Yes if X and Y are on the same side of AB (Theorem 10: A, B, X, Y are concyclic). If they are on opposite sides, not in general.
ABCD is a cyclic quadrilateral with ∠D = 100° and x = ∠B (opposite to D). Opposite angles add up to 180°: x = 180° − 100° = 80°.
∠A + ∠C = 80° + 100° = 180° and ∠B + ∠D = 110° + 70° = 180°. Opposite angles add up to 180°, so by Theorem 12 such a quadrilateral can be cyclic; yes, it can be drawn.
Half chord = √(169 − 25) = 12. Chord = 24 cm.
Half of 70°: 35°.
r = 13, half chord = 12. d = √(169 − 144) = 5 cm.
Half chord = √(225 − 81) = 12. Chord = 24 cm.
Let AB be a chord and O the centre. OA = OB (radii), so O is equidistant from A and B and therefore lies on the perpendicular bisector of AB.
90°. The arc AB (the half circle not containing C) subtends 180° at the centre, so ∠ACB = ½ × 180° = 90°.
∠C = 180° − 75° = 105°. ∠D = 180° − 110° = 70°.
∠P + ∠R = 180°: 5x − 10 = 180, so x = 38.
∠P = 2(38) + 10 = 86° and ∠R = 3(38) − 20 = 94°.
r² = 8² + 6² = 100, so r = 10 cm.
If the sides are in the order 5, 5, 12, 12 (a kite), the diagonal joining the two vertices where unequal sides meet is a diameter: √(5² + 12²) = 13. The two triangles are right-angled (angle in a semicircle) with legs 5 and 12: area = 2 × ½ × 5 × 12 = 60 square units.
If the order is 5, 12, 5, 12, it is a cyclic parallelogram, hence a 5 × 12 rectangle with area 60 too. (Brahmagupta’s formula √((s−a)(s−b)(s−c)(s−d)) with s = 17 also gives √(12 · 12 · 5 · 5) = 60.)
For a side AB, the centre is on the same side of AB as the other vertices exactly when the inscribed angle that AB subtends at those vertices is acute (the centre is then on the major-arc side); it is on AB when that angle is 90° (AB a diameter), and on the other side when it is obtuse.
So: find the angle each of the four sides subtends at an opposite vertex (the angle between a side and a diagonal, e.g. ∠ACB for side AB). The centre is inside iff all four are acute; on a side if one is 90°; outside if one is obtuse.
Let chords AB and CD of equal length meet at P. Let OM ⟂ AB and ON ⟂ CD (M, N are midpoints). Equal chords are equidistant: OM = ON. In right triangles OMP and ONP: OP is common and OM = ON, so by RHS PM = PN.
Also AM = MB = CN = ND (halves of equal chords). The two segments of AB are AM + MP and AM − MP (or MB ± MP), i.e. a + m and a − m where a = AM, m = PM; the same two values arise for CD. Hence the segments of one chord equal the corresponding segments of the other (label so PA = PC and PB = PD).
The chord and the distance give r = √(9 + 9) = 3√2 ≈ 4.24 cm. Draw AB = 6 cm; draw its perpendicular bisector meeting AB at M; mark O on it with OM = 3 cm; draw the circle with centre O and radius OA.
(Hint construction: ∠AOB = 90°, so the angle at the circumference is 45°. Draw ΔABP with AB = 6 and ∠P = 45° (base angles 67.5°); its circumcircle is the required circle.)
In a parallelogram opposite angles are equal. In a cyclic quadrilateral they add up to 180°. So each of them is 90°, and a parallelogram with a right angle is a rectangle. Conversely every rectangle is cyclic.
Each angle of the rectangle is 90°, so each diagonal subtends 90° at the opposite vertices, and by Theorem 9 subtends 180° at the centre: each diagonal is a diameter. Two diameters meet at the centre, so the diagonals intersect at the centre.
A chord of length l is at distance d = √(r² − l²/4) from the centre, and its midpoint is the foot of the perpendicular. So all midpoints are at distance d from O: they form a circle concentric with the given circle, of radius d.
Equal chords are equidistant from the centre, so O is at the same distance from the lines AB and AC, which means it lies on the bisector of ∠BAC. (Also, ΔOAB ≅ ΔOAC by SSS: OA common, OB = OC, AB = AC, so ∠OAB = ∠OAC.)
Distances: √(r² − 25) for the 10 cm chord and √(r² − 144) for the 24 cm chord. Their difference is 7: √(r² − 25) − √(r² − 144) = 7.
Try r = 13: √144 − √25 = 12 − 5 = 7 ✓. So r = 13 cm. (Squaring gives the same unique positive solution.)
Each side subtends 360°/6 = 60° at the centre, so the triangle formed with two radii is equilateral: side = r.
Distance of a side from the centre = √(r² − (r/2)²) = (√3/2) r.
N and O lie on the same side of chord MP, so the angles they subtend are in the same segment: ∠MOP = ∠MNP. Also, since MN is a diameter, ∠MPN = ∠MON = 90° (angle in a semicircle).
∠CDE = 180° − ∠CDA (linear pair) and ∠ABC = 180° − ∠CDA (opposite angles of a cyclic quadrilateral). So ∠CDE = ∠ABC.
For a chord AB, in ΔOAB: AB ≤ OA + OB = 2r (triangle inequality). Equality holds only when O lies on AB, i.e. AB is a diameter. Or: l = 2√(r² − d²) ≤ 2r, with equality when d = 0.
A chord through A at distance d from O has length 2√(r² − d²), which gets smaller as d grows. The distance from O to any line through A is at most OA, with the maximum OA exactly when the line is perpendicular to OA. So the shortest chord has d = OA, i.e. it is perpendicular to OA.
OA = OP = OQ (radii) give two isosceles triangles with base angles a and b. The angles of the big triangle: a + b + (a + b) = 180°, so a + b = 90°. The angle at A is a + b = 90°.
Reflection in the diameter AB maps the circle to itself. Since CC′ ⟂ AB and DD′ ⟂ AB, C and C′ are mirror images, and so are D and D′. Hence the chord CD maps to C′D′, and the midpoint M maps to M′. A point and its mirror image are joined by a line perpendicular to the mirror line: MM′ ⟂ AB.
Join O to A, B, C, D. OA = OB = OC = OD, so ΔOAB, ΔOBC, ΔOCD, ΔODA are isosceles. Call the base angles p, q for ΔOAB (p = q), x for ΔOBC, u for ΔOCD (u = v) and y for ΔODA.
∠A = p + y, ∠B = q + x, ∠C = x + u, ∠D = u + y. Then ∠A + ∠C = p + y + x + u and ∠B + ∠D = p + x + u + y (as q = p). So the two sums are equal; they add up to 360°, so each is 180°.