Let the abscissae of the two points $P$ and $Q$ be the roots of $2x^2 – rx + p = 0$ and the ordinates of $P$ and $Q$ be the roots of $x^2 – sx – q = 0$ . If the equation of the circle described on $PQ$ as diameter is $2(x^2 + y^2) – 11x – 14y – 22 = 0$ , then $2r + s – 2q + p$ is equal to _________.

Question

Let the abscissae of the two points  $P$  and  $Q$  be the roots of  $2x^2 – rx + p = 0$  and the ordinates of  $P$  and  $Q$  be the roots of  $x^2 – sx – q = 0$ . If the equation of the circle described on  $PQ$  as diameter is  $2(x^2 + y^2) – 11x – 14y – 22 = 0$ , then  $2r + s – 2q + p$  is equal to _________.

cdquestions Admin · Accepted Answer

Suppose P(x 1 , y 1 ) & Q(x 2 , y 2 ) Solve: 2x 2  – rx + p = 0 for x 1  and x 2 Solve: x 2  – sx – q = 0 for y 1  and y 2 So, the quation of circle = (x – x 1 ) (x – x 2 ) + (y – y 1 ) (y – y 2 ) = 0 x 2  – (x 1  + x 2 )x + x 1 x 2  + y 2  – (y 1  + y 2 )y + y 1 y 2  = 0 x2− $\frac r2$ x+ $\frac p2$ +y 2 +sy−q = 0 2x 2  + 2y 2  – rx + 2sy + p – 2q = 0 On comparing with 2x 2  + 2y 2  – 11x – 14y – 22 = 0 r = 11, s = 7, p – 2q = –22 Then, 2r + s + p – 2q = 22 + 7 – 22 = 7 So, the answer is 7.

Correct Answer: 7

Solution and Explanation

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Concepts Used:

Introduction to Three Dimensional Geometry

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