P and q are centers of circles of radii 9 and 2 respectively pq is 7 cm r is the centre of the circle of radious x which touches the above circle externally given that angle prq is 90 write the equation in x and solve it
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In right triangle PQR, by Pythagoras theorem
PQ2
= PR2
+PQ2
⇒ 17²= (x+9)² + (x+2)²
⇒ x² + 11x – 102 = 0
⇒ x² + 17x – 6x – 102 = 0
⇒ x(x+17) – 6(x+17) = 0
⇒ (x-6)(x+17) = 0
⇒ x = 6 or x = -17
⇒ x = 6 cm (x can’t be negative)
⇒ 17²= (x+9)² + (x+2)²
⇒ x² + 11x – 102 = 0
⇒ x² + 17x – 6x – 102 = 0
⇒ x(x+17) – 6(x+17) = 0
⇒ (x-6)(x+17) = 0
⇒ x = 6 or x = -17
⇒ x = 6 cm (x can’t be negative)
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