Q15 of 31 Page 1

P & Q are centres of circles of radii 9 cm and 2 cm respectively. PQ= 17cm. R is the centre of the circle of radius x cm which touches the above circle externally. Given that angle PRQ is 90o. Write an equation in x and solve it.


Let’s make a figure for the above problem


Given: P & Q are centres of circles of radii 9 cm and 2 cm respectively. PQ= 17cm. R is the centre of the circle of radius x cm which touches the above circle externally and ∠PRQ = 90°


To find: Equation in x, and value of x


In ΔPRQ, ∠PRQ = 90° [Given]


PR = (9 + x) cm


QR = (2 + x) cm


PQ = 17 cm [Given]


Also, By Pythagoras theorem [Hypotenuse2 = Base2 + Perpendicular2]


PQ2 = PR2 + QR2


⇒ 172 = (9 + x)2 + (2 + x)2


⇒ 289 = 81 + 18x + x2 + 4 + 4x + x2


⇒ 289 = 85 + 18x + 2x2


⇒ 2x2 + 22x – 204 = 0


⇒ x2 + 11x – 102 = 0


⇒ x2 + 17x – 6x – 102 = 0


⇒ x(x + 17) – 6(x + 17) = 0


⇒ x = 6 cm or –17cm


As negative distance is not possible


⇒ x = 6 cm


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