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Class 10 Mathematics | Chapter 10 : Circles | Full Exercise Solutions
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Exercise 10.1
Q. How many tangents can a circle have?
A circle has infinitely many tangents, because there is one tangent at every point on the circle, and a circle has infinitely many points.
Q. Fill in the blanks:
(i) A tangent to a circle intersects it in one point.
(ii) A line intersecting a circle in two points is called a secant.
(iii) A circle can have two parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called the point of contact.
Q. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is: (A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 cm.
Solution:
Since PQ is a tangent at P, OP ⊥ PQ (tangent ⊥ radius at point of contact).
In right triangle OPQ: OQ² = OP² + PQ²
12² = 5² + PQ² ⇒ 144 = 25 + PQ² ⇒ PQ² = 119
PQ = √119 cm.
Correct option: (D)
Q. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
Draw any circle with centre O and a line l outside/away from it.
Draw a line m parallel to l, at a distance equal to the radius, touching the circle at exactly one point — this is the required tangent.
Draw another line n parallel to l, passing through the interior of the circle so it cuts the circle at two points — this is the required secant.
(Best drawn by hand/geometry box; the key idea is choosing the perpendicular distance from the centre equal to the radius for the tangent, and less than the radius for the secant.)
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Exercise 10.2
Q. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is: (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm.
Solution:
Let the radius be r. Since the tangent is perpendicular to the radius at the point of contact, we get a right triangle with hypotenuse OQ = 25 cm and one leg = tangent length = 24 cm.
r² = OQ² − (tangent)² = 25² − 24² = 625 − 576 = 49
r = 7 cm.
Correct option: (A)
Q. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to: (A) 60° (B) 70° (C) 80° (D) 90°.
Solution:
OPTQ is a quadrilateral. Since the tangent is perpendicular to the radius at the point of contact, ∠OPT = ∠OQT = 90°.
Sum of angles of quadrilateral OPTQ = 360°
∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360°
110° + 90° + ∠PTQ + 90° = 360°
∠PTQ = 360° − 290° = 70°
Correct option: (B)
Q. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to: (A) 50° (B) 60° (C) 70° (D) 80°.
Solution:
Since PA = PB (tangents from an external point are equal), OP bisects ∠APB, and also OP bisects the angle ∠AOB at the centre by symmetry of the two congruent right triangles OAP and OBP.
In right triangle OAP, ∠OAP = 90°, and ∠APO = half of ∠APB = 40°.
∠POA = 180° − 90° − 40° = 50°
Correct option: (A)
Q. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Let AB be a diameter of a circle with centre O, and let l and m be the tangents at A and B respectively.
By the tangent–radius property, l ⊥ OA and m ⊥ OB.
Since A, O, B are collinear (AB is a diameter, a straight line through the centre), OA and OB lie along the same line.
Two lines (l and m) that are each perpendicular to the same line are parallel to each other.
Hence, l ∥ m, i.e., the tangents at the two ends of a diameter are parallel. (Q.E.D.)
Q. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
Let XY be a tangent to a circle with centre O, touching the circle at point P.
Suppose, for contradiction, that the perpendicular to XY at P does not pass through O; call this perpendicular line PZ.
We already know (Theorem: tangent ⊥ radius) that OP is perpendicular to XY at P.
If PZ ≠ OP but both are perpendicular to XY at the same point P, then two distinct lines would be perpendicular to XY at P, which is impossible (only one perpendicular can be drawn to a line at a given point).
Hence PZ must coincide with OP, so the perpendicular at P passes through the centre O. (Q.E.D.)
Q. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
Let the radius be r and O be the centre. Since tangent ⊥ radius at the point of contact, we get a right triangle with hypotenuse OA = 5 cm and one leg = 4 cm (tangent length).
r² = OA² − (tangent)² = 5² − 4² = 25 − 16 = 9
r = 3 cm.
Q. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Let O be the common centre, and let the chord AB of the larger circle touch the smaller circle at P.
Since AB is tangent to the smaller circle at P, OP ⊥ AB, and OP = 3 cm (radius of smaller circle), OA = 5 cm (radius of larger circle).
In right triangle OPA: AP² = OA² − OP² = 25 − 9 = 16 ⇒ AP = 4 cm
Since the perpendicular from the centre bisects the chord, AB = 2 × AP = 8 cm.
Q. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC.
Solution:
Let the circle touch sides AB, BC, CD and DA at points P, Q, R and S respectively.
Using the property that tangents drawn from an external point to a circle are equal in length:
AP = AS (tangents from A)
BP = BQ (tangents from B)
CR = CQ (tangents from C)
DR = DS (tangents from D)
Adding these four equations:
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC (Q.E.D.)
Q. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O, and another tangent AB with point of contact C intersects XY at A and X′Y′ at B. Prove that ∠AOB = 90°.
Solution:
Join OC. Since AP and AC are tangents from external point A (P being the point where XY touches the circle), AP = AC, and OA bisects ∠PAC.
Similarly, OB bisects ∠QBC (Q being the point where X′Y′ touches the circle), since BQ = BC.
Let ∠PAC = 2α and ∠QBC = 2β. Then ∠OAC = α and ∠OBC = β.
Since XY ∥ X′Y′, the co-interior angles at A and B (∠PAB and ∠QBA) are supplementary:
2α + 2β = 180° ⇒ α + β = 90°
In triangle AOB: ∠OAB + ∠OBA + ∠AOB = 180°
α + β + ∠AOB = 180° ⇒ 90° + ∠AOB = 180° ⇒ ∠AOB = 90° (Q.E.D.)
Q. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution:
Let PA and PB be two tangents drawn from an external point P, touching the circle at A and B, with centre O.
Since PA ⊥ OA and PB ⊥ OB (tangent ⊥ radius), ∠OAP = ∠OBP = 90°.
In quadrilateral OAPB, the sum of all interior angles is 360°:
∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
∠AOB + 90° + ∠APB + 90° = 360°
∠AOB + ∠APB = 180°
Hence ∠APB (angle between the tangents) and ∠AOB (angle subtended at the centre) are supplementary. (Q.E.D.)
Q. Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Let ABCD be a parallelogram circumscribing a circle that touches sides AB, BC, CD, DA at P, Q, R, S respectively.
Using equal tangent lengths from each vertex:
AP = AS, BP = BQ, CR = CQ, DR = DS
Adding: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Since ABCD is a parallelogram, AB = CD and AD = BC (opposite sides equal).
Substituting: AB + AB = AD + AD ⇒ 2AB = 2AD ⇒ AB = AD
So all four sides AB = BC = CD = DA are equal, meaning the parallelogram is a rhombus. (Q.E.D.)
Q. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
Solution:
Let the circle touch AB, BC, CA at F, D, E respectively. Let AF = AE = x (tangents from A are equal).
Given BD = BF = 8 cm and DC = CE = 6 cm (tangents from B and C respectively).
So AB = x + 8, AC = x + 6, BC = 8 + 6 = 14 cm.
Semi-perimeter, s = (AB + BC + CA)/2 = (x + 8 + 14 + x + 6)/2 = x + 14
Using Heron’s formula, Area = √[s(s−a)(s−b)(s−c)] where a = BC = 14, b = CA = x+6, c = AB = x+8:
s − a = x, s − b = 8, s − c = 6
Area = √[(x+14) · x · 8 · 6] = √[48x(x+14)]
Also, Area = r × s = 4(x + 14), since the incircle radius r = 4 cm.
Equating: 4(x+14) = √[48x(x+14)]
Squaring both sides: 16(x+14)² = 48x(x+14)
Dividing both sides by (x+14): 16(x+14) = 48x
16x + 224 = 48x ⇒ 224 = 32x ⇒ x = 7
Therefore, AB = x + 8 = 15 cm and AC = x + 6 = 13 cm.
Q. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Let ABCD be a quadrilateral circumscribing a circle with centre O, touching sides AB, BC, CD, DA at P, Q, R, S respectively. Join O to P, Q, R, S.
Since tangents from an external point are equal and the line from the centre to an external point bisects the angle between the two tangents from that point, OA bisects ∠SAP, OB bisects ∠PBQ, OC bisects ∠QCR, and OD bisects ∠RDS.
Also, in triangles formed by O with each pair of adjacent tangent points, the two base angles are equal (isosceles triangles, since OP = OQ = OR = OS = radius, and each tangent-radius angle is 90°).
Let ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7, ∠8 denote consecutive angles around O formed by joining O to P, Q, R, S (going around the circle): ∠1 = ∠2 (at vertex A), ∠3 = ∠4 (at vertex B), ∠5 = ∠6 (at vertex C), ∠7 = ∠8 (at vertex D).
The sum of all eight angles around point O is 360°:
(∠1+∠2) + (∠3+∠4) + (∠5+∠6) + (∠7+∠8) = 360°
Grouping as ∠AOB (=∠2+∠3), ∠BOC (=∠4+∠5), ∠COD (=∠6+∠7), ∠DOA (=∠8+∠1), and using the equal-angle pairs, it follows that:
∠AOB + ∠COD = 180° and ∠BOC + ∠DOA = 180°
Hence, the opposite sides AB & CD, and BC & DA, subtend supplementary angles at the centre O. (Q.E.D.)
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