Question

Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.

Answer 1

According to the question,

Two circles with centers O and O’ of radii 3 cm and 4 cm, respectively, intersect at two points P and Q, such that OP and O’P are tangents to the two circles and PQ is a common chord.

To Find: Length of common chord PQ

∠OPO’ = 90°  ...[Tangent at a point on the circle is perpendicular to the radius through point of contact]

So OPO is a right-angled triangle at P

Using Pythagoras in ΔOPO’, we have

(OO’)² = (O’P)² + (OP)²

(OO’)² = (4)² + (3)²

(OO’)² = 25

OO’ = 5 cm

Let ON = x cm and NO’ = 5 – x cm

In right angled triangle ONP

(ON)² + (PN)²= (OP)²

x² + (PN)² = (3)²

(PN)² = 9 – x²   ...[1]

In right angled triangle O’NP

(O’N)² + (PN)² = (O’P)²

(5 – x)² + (PN)² = (4)²

25 – 10x + x² + (PN)² = 16

(PN)² = – x²+ 10x – 9  ...[2]

From [1] and [2]

9 – x² = – x² + 10x – 9

10x = 18

x = 1.8

From (1) we have

(PN)² = 9 – (1.8)²

= 9 – 3.24

= 5.76

PN = 2.4 cm

PQ = 2PN

= 2(2.4)

= 4.8 cm