Mathematics
P and Q are centers of a circle of radii 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm which touches the above circles externally. Given that ∠PRQ = 90°, write an equation in x and solve it.
Answer
From figure,
In △PQR,
By pythagoras theorem,
⇒ PQ2 = PR2 + QR2
⇒ 172 = (x + 9)2 + (x + 2)2
⇒ 289 = x2 + 81 + 18x + x2 + 4 + 4x
⇒ 289 = 2x2 + 22x + 85
⇒ 2x2 + 22x + 85 - 289 = 0
⇒ 2x2 + 22x - 204 = 0
⇒ 2(x2 + 11x - 102) = 0
⇒ x2 + 11x - 102 = 0
⇒ x2 + 17x - 6x - 102 = 0
⇒ x(x + 17) - 6(x + 17) = 0
⇒ (x - 6)(x + 17) = 0
⇒ x - 6 = 0 or x + 17 = 0
⇒ x = 6 or x = -17.
Since, radius cannot be negative.
⇒ x = 6 cm.
Hence, equation is x2 + 11x - 102 = 0 and x = 6 cm.
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