Question

P and Q are the centre of circles of radius 9 cm and 2 cm respectively; PQ = 17 cm. R is the centre of circle of radius x cm, which touches the above circles externally, given that ∠ PRQ = 90°. Write an equation in x and solve it.

Answer 1

Let the circle with centre R touch the given two circles at A and B. Then, P, A, R are collinear and Q, B, R are collinear.

Since, ∠PRQ = 90°,
by Pythagoras theorem,
PQ² = PR² + QR²
⇒ 17² = (9 + x)² + (2 + x)²
⇒ x² + 11x - 102 = 0
⇒ (x + 17)(x - 6) = 0
⇒ x = 6 cm            ....( x = - 17 is not possible ).