# Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q

**Solution:**

**Steps of construction:**

- Draw a circle with O as centre and radius as 3 cm.
- Draw a diameter of it and extend both the sides and mark the points as P, Q such that OP = OQ = 7 cm.
- Draw the perpendicular bisectors of OP and OQ to intersect PQ at M and N respectively.
- With M as centre and OM as radius draw a circle to cut the given circle at A and C. With N as the centre and ON as radius draw a circle to cut the given circle at B and D.
- Join PA, PC, QB, QD.

PA, PC and QB, QD are the required tangents from P and Q respectively.

Proof:

∠PAO = ∠QBO = 90° (Angle in a semi-circle)

∴ PA ⊥ AO, QB ⊥ BO

In right angle triangle PAO,

OP = OQ = 7cm (By construction)

OA = OB = 3cm (radius of the given circle)

PA² = (OP)² - (OA)²

= (7)² - (3)²

= 49 - 9

= 40

PA = √40

= 6.3 (approx.)

Also, in right triangle QBO,

QB² = (OQ)² - (OB)²

= (7)² - (3)²

= 49 - 9

= 40

QB = √40

= 6.3 (approx)

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 11

**Video Solution:**

## Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q

NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.2 Question 3

**Summary:**

A circle of radius 3 cm is drawn. Two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre has been taken and the tangents PA and QB each of length 6.3 cm has been constructed.

**☛ Related Questions:**

- Draw a pair of tangents to a circle of radius 5 cm which is inclined to each other at an angle of 60°.
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- Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is perpendicular to AC. The circle through B, C and D is drawn. Construct the tangents from A to this circle.
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