Chapter 10: Circles (NCERT Solutions)

Exercise 10.1: Tangents to a Circle

Q1. How many tangents can a circle have?

A circle can have infinitely many tangents. Each point on the circle is a point of tangency for exactly one tangent.

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.

Q3. 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

OP ⊥ PQ (radius ⊥ tangent).
In right ΔOPQ: OQ² = OP² + PQ².
12² = 5² + PQ².
144 = 25 + PQ².
PQ² = 119.
PQ = √119 cm (D).

Q4. 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.

[Construction required - Draw a circle, a tangent line touching at one point, and a parallel secant cutting at two points]

Exercise 10.2: Number of Tangents from a Point

Q1. 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

OQ² = r² + tangent².
25² = r² + 24².
625 = r² + 576.
r² = 49.
r = 7 cm (A).

Q2. 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°

∠OPT = ∠OQT = 90° (radius ⊥ tangent).
In quadrilateral OPTQ:
∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360°.
110° + 90° + ∠PTQ + 90° = 360°.
∠PTQ = 360° - 290° = 70° (B).

Q3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to: (A) 50° (B) 60° (C) 70° (D) 80°

∠APB = 80°.
In ΔOAP: ∠OAP = 90°.
∠OPA = (180° - 80°)/2 = 50° (tangents from P are equal, so isosceles).
∠POA = 180° - 90° - 50° = 50° (A).

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Let AB be diameter with tangents at A and B.
Radius OA ⊥ tangent at A ⇒ ∠ = 90°.
Radius OB ⊥ tangent at B ⇒ ∠ = 90°.
Both tangents make 90° with diameter AB.
Hence, tangents are parallel.

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Let P be point of contact, tangent be l.
Radius OP is perpendicular to tangent l.
Any perpendicular to l at P will be along OP (unique perpendicular).
Hence, it passes through centre O.

Q6. 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.

OA² = r² + tangent².
5² = r² + 4².
25 = r² + 16.
r² = 9.
r = 3 cm.

Q7. 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.

Chord AB of larger circle is tangent to smaller circle at P.
OP ⊥ AB (perpendicular from centre to chord).
OP = 3 cm (radius of smaller circle).
OA = 5 cm (radius of larger circle).
In right ΔOPA: AP² = OA² - OP² = 25 - 9 = 16.
AP = 4 cm.
AB = 2 × AP = 8 cm.

Q8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC.

Tangents from external point are equal.
From A: AP = AS.
From B: BP = BQ.
From C: CQ = CR.
From D: DR = DS.
AB + CD = (AP + BP) + (CR + DR).
= (AS + BQ) + (CQ + DS).
= (AS + DS) + (BQ + CQ).
= AD + BC. Proved.

Q9. 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 intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.

Let XY touch at P, X'Y' at Q.
PA = CA (tangents from A).
QB = CB (tangents from B).
OP ⊥ XY, OQ ⊥ X'Y'.
Since XY || X'Y', OP and OQ are in opposite directions, i.e., POQ is diameter.
∠PCQ = 90° (angle in semicircle on diameter PQ).
In ΔPAC and ΔCAO: PA = CA, ∠OPA = ∠OCA = 90°.
∴ ∠PAO = ∠CAO.
Similarly, ∠QBO = ∠CBO.
In ΔAOB: ∠OAB + ∠OBA = (∠PAO + ∠CAO) + (∠QBO + ∠CBO) = 90° (as PQ is straight).
∴ ∠AOB = 90°. Proved.

Q10. 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.

Let P be external point, PA and PB be tangents.
∠OAP = ∠OBP = 90°.
In quadrilateral OAPB:
∠AOB + ∠OAP + ∠APB + ∠OBP = 360°.
∠AOB + 90° + ∠APB + 90° = 360°.
∠AOB + ∠APB = 180°.
Hence, they are supplementary. Proved.

Q11. Prove that the parallelogram circumscribing a circle is a rhombus.

Let ABCD be parallelogram circumscribing circle.
AB + CD = AD + BC (tangent property from Q8).
But AB = CD and AD = BC (opposite sides of parallelogram).
∴ 2AB = 2AD.
∴ AB = AD.
Since all sides are equal, ABCD is a rhombus. Proved.

Q12. 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.

Let circle touch AB at E, AC at F.
BD = BE = 8 cm (tangents from B).
CD = CF = 6 cm (tangents from C).
Let AE = AF = x cm (tangents from A).
AB = AE + EB = x + 8.
AC = AF + FC = x + 6.
BC = BD + DC = 8 + 6 = 14 cm.
Semi-perimeter s = (AB + BC + AC)/2 = (x+8+14+x+6)/2 = (2x+28)/2 = x+14.
Area = √[s(s-AB)(s-BC)(s-AC)].
Also, Area = rs = 4(x+14).
s-AB = x+14-x-8 = 6.
s-BC = x+14-14 = x.
s-AC = x+14-x-6 = 8.
√[(x+14)(6)(x)(8)] = 4(x+14).
√[48x(x+14)] = 4(x+14).
48x(x+14) = 16(x+14)².
3x = (x+14).
3x = x + 14.
2x = 14.
x = 7 cm.
AB = 7 + 8 = 15 cm.
AC = 7 + 6 = 13 cm.

Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Let ABCD circumscribe circle with center O, touching at P, Q, R, S.
∠AOP = ∠AOS (OP and OS are equal radii, tangents from A).
Similarly for other vertices.
Sum of all angles at O = 360°.
2(∠AOP + ∠BOP + ∠COQ + ∠DOQ) = 360°.
Grouping: ∠AOB + ∠COD = 180°.
Similarly, ∠BOC + ∠DOA = 180°.
Hence, opposite sides subtend supplementary angles. Proved.

Circles - RD Sharma Important Questions

Q1. If from an external point B, two tangents BP and BQ are drawn to a circle with centre O, prove that BO is the perpendicular bisector of PQ.

BP = BQ (tangents from external point).
OP = OQ (radii).
BO = BO (common).
ΔBOP ≅ ΔBOQ (SSS).
∴ ∠BOP = ∠BOQ.
In ΔPOM and ΔQOM (M is intersection):
OP = OQ, OM common, ∠POM = ∠QOM.
ΔPOM ≅ ΔQOM.
∴ PM = QM and ∠PMO = ∠QMO = 90°.
Hence, BO is perpendicular bisector of PQ. Proved.

Q2. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

∠OPT = 90° (radius ⊥ tangent).
In ΔOPQ: OP = OQ (radii), so isosceles.
Let ∠OPQ = ∠OQP = θ.
∠POQ = 180° - 2θ.
∠PTQ = 180° - ∠POQ = 180° - (180° - 2θ) = 2θ.
∴ ∠PTQ = 2∠OPQ. Proved.

Q3. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

Draw OM ⊥ PQ. M is midpoint.
PM = 4 cm.
OM² = OP² - PM² = 25 - 16 = 9.
OM = 3 cm.
Let TP = TQ = x (tangents from T).
In ΔTMP: TM² = TP² - PM² = x² - 16.
Also, ∠OPT = 90°.
In ΔOPT: OT² = OP² + TP² = 25 + x².
OT = OM + MT.
Using similar triangles or direct calculation:
From ΔOPT: tan(∠POT) = x/5.
From geometry: TM/OM = TP/OP.
Solving: TP = 20/3 cm.

Q4. Prove that tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Let AB be chord, tangents at A and B meet at T.
TA = TB (tangents from T).
∴ ΔTAB is isosceles.
∴ ∠TAB = ∠TBA.
These are angles made by tangents with chord. Proved.

Q5. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.

This is NCERT Q8 from Exercise 10.2. Standard tangent property proves AB + CD = AD + BC.

Q6. In Fig., AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to what?

∠ABC = 90° (angle in semicircle).
∠BAC = 180° - 90° - 50° = 40°.
∠OAT = 90° (radius ⊥ tangent).
∠OAB = ∠BAC = 40°.
∠BAT = 90° - 40° = 50°.

Q7. Two circles touch internally. The sum of their areas is 116π cm² and distance between their centres is 6 cm. Find the radii.

Let radii be R and r (R > r).
R - r = 6 (internal touch).
πR² + πr² = 116π.
R² + r² = 116.
R = r + 6.
(r+6)² + r² = 116.
r² + 12r + 36 + r² = 116.
2r² + 12r - 80 = 0.
r² + 6r - 40 = 0.
(r+10)(r-4) = 0.
r = 4 cm (taking positive).
R = 10 cm.
Radii: 10 cm and 4 cm.

Q8. If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by r = (a+b-c)/2.

Area = rs where s = (a+b+c)/2.
Also, Area = ab/2.
ab/2 = r(a+b+c)/2.
ab = r(a+b+c).
r = ab/(a+b+c).
Multiply numerator and denominator by (a+b-c):
r = ab(a+b-c)/[(a+b+c)(a+b-c)].
= ab(a+b-c)/[(a+b)² - c²].
= ab(a+b-c)/(a² + b² + 2ab - c²).
Since c² = a² + b²:
= ab(a+b-c)/(2ab).
= (a+b-c)/2. Proved.

Q9-Q20. [Additional problems on: Common tangents to two circles, tangent length calculations, chord properties, inscribed and circumscribed figures]

Key concepts used:
• Tangent ⊥ radius at point of contact
• Tangents from external point are equal
• Angle between tangent and chord = inscribed angle in alternate segment
• Properties of cyclic quadrilaterals and tangent circles

Circles - Formulas & PYQs

Key Theorems & Formulas

1. Tangent to a Circle

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

If OP is radius and PT is tangent at P, then OP ⊥ PT.

2. Tangents from External Point

The lengths of tangents drawn from an external point to a circle are equal.

If PA and PB are tangents from P, then PA = PB.

3. Length of Tangent Formula

If P is external point at distance d from centre O, and r is radius:

Length of tangent = √(d² - r²)

4. Angle between Tangents

If two tangents PA and PB are drawn from P:

∠APB + ∠AOB = 180° (supplementary)

5. Quadrilateral Circumscribing Circle

AB + CD = BC + DA

(Sum of opposite sides are equal)

6. Incircle of Right Triangle

For right triangle with sides a, b and hypotenuse c:

r = (a + b - c)/2

Previous Year Questions (CBSE/JKBOSE)

Q1. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (CBSE 2020)

Standard theorem proof using contradiction method.

Q2. In Fig., PA and PB are tangents to the circle with centre O. If ∠APB = 60°, find ∠OAB. (CBSE 2019)

∠AOB = 180° - 60° = 120°.
In ΔOAB: OA = OB (radii).
∠OAB = ∠OBA = (180° - 120°)/2 = 30°.

Q3. 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. (CBSE 2018)

NCERT Ex 10.2 Q7. Length = 8 cm.

Q4. Prove that the parallelogram circumscribing a circle is a rhombus. (CBSE 2017)

NCERT Ex 10.2 Q11. Standard proof using tangent properties.

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