# RD Sharma Solutions for Class 10 Maths Chapter 10 Circles MCQs and VSAQs

**VSAQs**

**Answer each of the following questions either in one word or one sentence or as per requirement of the questions :**

**1. **In the figure, PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q. If PB = 10 cm and CQ = 2 cm, what is the length PC ?

**Solution**

In the figure, PA and PB are the tangents to the circle drawn from P

CD is the third tangent to the circle drawn at Q

PB = 10 cm, CQ = 2 cm

PA and PB are tangents to the circle

PA = PB = 10 cm

Similarly CQ and CA are tangents to the circle

CQ = CA = 2 cm

PC = PA – CA = 10 – 2 = 8 cm

**2. What is the distance between two parallel tangents of a circle of radius 4 cm ?**

**Solution**

TT’ and SS’ are two tangents of a circle with centre O and radius 4 cm and TT’ || SS’

OP and OQ are joined

Now OP is the radius and TPT’ is the tangent

OP ⊥ TPT’

Similar OQ ⊥ SS’

But TT’ || SS’

POQ is the diameter

Which is 4×2 = 8 cm

Distance between the two parallel tangents is 8 cm

**3. The length of tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. What is the radius of the circle ?**

**Solution**

PA is a tangent to the circle from P at a distance of 5 cm from the centre O

PA = 4 cm

OA is joined and let OA = r

Now in right ∆OAP,

OP² = OA² + PA²

⇒ (5)² = r² + (4)²

⇒ 25 = r + 16

⇒ r² = 25 – 16 = 9 = (3)²

r = 3

Radius of the circle = 3 cm

**4. Two tangents TP and TQ are drawn from an external point T to a circle with centre O as shown in the following figure. If they are inclined to each other at an angle of 100°, then what is the value of ∠POQ ?**

**Solution**

TP and TQ are the tangents from T to the circle with centre O and ∠PTQ = 100°

OT, OP and OQ are joined

OP and OQ are radius

OP ⊥ PT and OQ ⊥ QT

Now in quadrilateral OPTQ,

∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360° **(Sum of angles of a quadrilateral)**

⇒ ∠POQ + 90° + 100° + 90° = 360°

⇒ ∠POQ + 280° = 360°

⇒ ∠POQ = 360° – 280° = 80°

Hence ∠POQ = 80°

**5. What is the distance between two parallel tangents to a circle of radius 5 cm?**

**Solution**

In a circle, the radius is 5 cm and centre is O

TT’ and SS’ are two tangents at P and Q to the circle

Such that TT’ || SS’

Join OP and OQ

OP is radius and TPT’ is the tangent

OP ⊥ TT’

Similarly OQ ⊥ SS’

POQ is the diameter of the circle

Now length of PQ = OP + OQ = 5 + 5 = 10 cm

Hence, distance between the two parallel tangents = 10 cm

**6. In Q. No. 1, if PB = 10 cm, what is the perimeter of ∆PCD ?**

**Solution**

In the figure, PB = 10 cm, CQ = 2 cm

PA and PB are tangents to the give from P

PA = PB = 10 cm

Similarly, CA and CQ are the tangents

CA = CQ = 2 cm

and DB and DQ are the tangents

DB = DQ

Now, perimeter of ∆PCD

PC + PD + CQ + DQ

= PC + CQ + PD + DQ

= PC + CA + PD + DB **{CQ = CA and DQ = DB}**

= PA + PB = 10 + 10 = 20 cm