Chapter 9: Circles

Q1Written answer

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

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Worked solution

From the figure,

Chord AB = 8 cm

OC is perpendicular to the chord AB

AC = CB = 4 cm

In right triangle OCA

OC² + CA² = OA²

OC² = 5²– 4² = 25 – 16 = 9

OC = 3 cm

Q2Written answer

Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

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Worked solution

NCERT Exemplar Class 10 Maths Chapter 9 Ex. 9.3 Question 1

We know that,

Radius ⊥ Tangent = OR ⊥ PR

i.e., ∠ORP = 90°

Likewise,

Radius ⊥ Tangent = OQ ⊥PQ

∠OQP = 90°

In quadrilateral ORPQ,

Sum of all interior angles = 360º

∠ORP + ∠RPQ+ ∠PQO + ∠QOR = 360º

90º + ∠RPQ + 90º + ∠QOR = 360º

Hence, ∠O + ∠P = 180^o

PROQ is a cyclic quadrilateral.

Q3Written answer

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that angle DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC.

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Worked solution

According to the question,

By RHS rule,

ΔOBC and ΔOBD are congruent

By CPCT

NCERT Exemplar Class 10 Maths Chapter 9 Ex. 9.3 Question 3

∠OBC and ∠ OBD are equal

Therefore,

∠OBC = ∠OBD =60°

In triangle OBC,

cos 60°=BC/OB

½ =BC/OB

OB=2BC

Hence proved

Q4Written answer

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

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Worked solution

Let the lines be l₁and l_2.

Assume that O touches l₁ and l₂ at M and N,

We get,

OM = ON (Radius of the circle)

Therefore,

From the centre ”O” of the circle, it has equal distance from l₁ & l₂.

In Δ OPM & OPN,

OM = ON (Radius of the circle)

∠OMP = ∠ONP (As, Radius is perpendicular to its tangent)

OP = OP (Common sides)

Therefore,

Δ OPM = ΔOPN (SSS congruence rule)

By C.P.C.T,

∠MPO = ∠NPO

So, l bisects ∠MPN.

Therefore, O lies on the bisector of the angle between l₁ & l₂.

Hence, we prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Q5Written answer

In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.

NCERT Exemplar Class 10 Maths Chapter 9 Ex. 9.3 Question 5

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Worked solution

According to the question,

AB = CD

NCERT Exemplar Class 10 Maths Chapter 9 Ex. 9.3 Question 5-1

Construction: Produce AB and CD, to intersect at P.

Proof:

Consider the circle with greater radius.

Tangents drawn from an external point to a circle are equal

AP = CP …(1)

Also,

Consider the circle with smaller radius.

Tangents drawn from an external point to a circle are equal

BP = BD …(2)

Subtract Equation (2) from (1). We Get

AP – BP = CP – BD

AB = CD

Hence Proved.

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Exercise 9.3