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

In ∆OO’P

(O’O)² = OP² + O’P²

= 3² + 4²

= 9 + 16

(OO’)² = 25

∴ OO’ = 5 cm

Since the line joining the centres of two intersecting circles is perpendicular bisector of their common chord.

OR ⊥ PQ and PR = RQ

Let OR be x, then O’R = 5 – x again Let PR = RQ = y cm

In ∆ORP,

OP² = OR² + PR²

9 = x² + y²  ...(1)

In ∆O’RP,

O’P² = O’R² + PR²

16 = (5 – x)² + y²

16 = 25 + x² – 10x + y²

16 = x² + y² + 25 – 10x

16 = 9 + 25 – 10x  ...[From (1)]

16 = 34 – 10x

10x = 34 – 16 = 18

x = `18/10` = 1.8 cm

Substitute the value of x = 1.8 in (1)

9 = (1.8)² + y²

y² = 9 – 3.24

y² = 5.76

y = `sqrt(5.76)` = 2.4 cm

Hence PQ = 2(2.4) = 4.8 cm

Length of the common chord PQ = 4.8 cm