NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 10.1
How many tangents can a circle have?
Answer
A circle can have infinitely many tangents. Every point on the circumference of a circle has exactly one tangent, and since a circle has infinite points on it, it has infinite tangents.
Q2. Fill in the blanks
(i) A tangent to a circle intersects it in one point(s).
(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 point of contact.
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. Find the length PQ.
Answer
OP ⊥ PQ (radius to tangent). In right ΔOPQ: PQ = √(OQ²–OP²) = √(144–25) = √119 = √119 cm ≈ 10.91 cm
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.
Answer
Draw a circle. Draw a line (l) outside the circle. Draw a line parallel to l that touches the circle at exactly one point — this is the tangent. Draw another line parallel to l that intersects the circle at two points — this is the secant.
Exercise 10.2
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. Find the radius of the circle.
Answer
Let radius = r. r² + 24² = 25²
⇒ r² = 625 – 576 = 49
⇒ r = 7 cm
In the figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to:
Answer
∠OPT = ∠OQT = 90° (radius ⊥ tangent). In quadrilateral OPTQ:
∠PTQ = 360° – 90° – 90° – 110° = 70°
If tangents PA and PB from an external point P to a circle with centre O are inclined to each other at 80°, then ∠POA is equal to:
Answer
∠APB = 80°. ∠AOB = 360° – 90° – 90° – 80° = 100°. PO bisects ∠AOB.
∠POA = 100°/2 = 50°
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer
Let AB be a diameter. Tangent at A is perpendicular to OA, and tangent at B is perpendicular to OB. Since OA and OB are along the same line (diameter), both tangents make 90° with the diameter. Two lines perpendicular to the same line are parallel. Hence proved.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer
We know that the radius through the point of contact is perpendicular to the tangent (theorem). There can be only one perpendicular to a line at a given point. Since the radius (which passes through the centre) is perpendicular to the tangent at the point of contact, the perpendicular at the point of contact must pass through the centre. Proved.
The length of a tangent from a point A at distance 5√2 cm from the centre of the circle is 5 cm. Find the radius of the circle.
Answer
r² = (5√2)² – 5² = 50 – 25 = 25
⇒ r = 5 cm
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.
Answer
The chord of the larger circle that is tangent to the smaller circle is perpendicular to the radius of the smaller circle at the point of tangency. Half-chord = √(5²–3²) = √16 = 4. Full chord = 2 × 4 = 8 cm
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Answer
Let the circle touch AB at P, BC at Q, CD at R, DA at S. Tangents from an external point are equal:
AP = AS, BP = BQ, CR = CQ, DR = DS.
AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC. Proved.
In the figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.
Answer
Join OC. In ΔOPA and ΔOCA: OP = OC (radii), OA = OA (common), AP = AC (tangents from A). So ΔOPA ≅ ΔOCA
⇒ ∠POA = ∠COA ... (1).
Similarly, ∠QOB = ∠COB ... (2).
∠POA + ∠COA + ∠COB + ∠QOB = 180° (PQ is diameter, straight line).
2∠COA + 2∠COB = 180°
⇒ ∠COA + ∠COB = 90°
⇒ ∠AOB = 90°. Proved.
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.
Answer
Let PA and PB be tangents from P, touching circle at A and B. ∠OAP = ∠OBP = 90°.
In quadrilateral OAPB: ∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
∠AOB + 90° + ∠APB + 90° = 360°
⇒ ∠AOB + ∠APB = 180° (supplementary). Proved.
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer
In parallelogram ABCD: AB = CD and BC = AD (opposite sides equal).
From Q8: AB + CD = AD + BC
⇒ 2AB = 2BC
⇒ AB = BC.
Since AB = BC = CD = AD, all sides are equal
⇒ ABCD is a rhombus. Proved.
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. Find the sides AB and AC.
Answer
Let circle touch AB at E and AC at F. BD = BE = 8 cm, CD = CF = 6 cm (tangents from same point). Let AE = AF = x.
AB = x + 8, AC = x + 6, BC = 14. Semi-perimeter s = (x+8+x+6+14)/2 = x + 14.
Area = r × s = 4(x+14). Also by Heron's: Area = √[s(s–a)(s–b)(s–c)] = √[(x+14)(x)(6)(8)] = √[48x(x+14)].
4(x+14) = √[48x(x+14)]
⇒ 16(x+14)² = 48x(x+14)
⇒ 16(x+14) = 48x
⇒ x+14 = 3x
⇒ x = 7.
AB = 15 cm, AC = 13 cm
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.
Answer
Let the circle touch AB at P, BC at Q, CD at R, DA at S. Join OP, OQ, OR, OS.
In triangles OAP and OAS: ∠1 = ∠2 (tangents from A are equal, and radii are equal
⇒ congruent triangles).
Similarly for all vertices. Sum of all angles at O = 360°.
∠AOB + ∠COD = (∠1+∠2) + (∠5+∠6) + ... = 180°. Similarly ∠BOC + ∠AOD = 180°. Proved.
Important Extra Questions
State and prove: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Answer
Statement: The tangent to a circle at any point is perpendicular to the radius drawn to that point.
Proof: Let O be the centre, P the point of contact, and l the tangent at P. Take any point Q on l other than P. Since l is a tangent, Q lies outside the circle
⇒ OQ > OP (OP is radius, Q is outside). This is true for every point Q on l except P. So OP is the shortest distance from O to l. The shortest distance from a point to a line is the perpendicular. Therefore, OP ⊥ l. Proved.
Prove that tangents drawn from an external point to a circle are equal in length.
Answer
Let PA and PB be tangents from external point P to circle with centre O, touching at A and B.
In ΔOAP and ΔOBP: OA = OB (radii), OP = OP (common), ∠OAP = ∠OBP = 90°.
By RHS congruence: ΔOAP ≅ ΔOBP
⇒ PA = PB. Proved.
Also, ∠OPA = ∠OPB (OP bisects the angle between tangents).