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

Given: OP = OQ = 4

O’P = O’Q = 3

OO’ is the perpendicular bisector of chord PQ.

Let R be the point of intersection of PQ and OO’.

Assume PR = QR = x and OR = y

In OPO’, OP² + O’P² = (OO’)² ⇒ OO’

= \(\sqrt{4^2+3^2}\) = 5

OR = y ⇒ OR = 5 – y

In ΔOPR, PR² + OR² = OP² ⇒ x² + y = 4² … (1)

In ΔO’PR, PR² + O’R² = O’P² ⇒ x² + (5 – y)² = 9 … (2)

(1) – (2) => y² – (25 + y² – 10y) = 16 – 9

⇒ y² – 25 – y² + 10y = 7 .

⇒ 10y = 25 + 7 ⇒ 10y =32

⇒ y = 3.2

Substituting y = 3.2 in (1), we get x = \(\sqrt{4^2+3.2^2}\)

x = 2.4

PQ = 2x ⇒ PQ = 4.8 cm