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प्रश्न
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.
उत्तर
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 ).
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