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 _________.

Answer 1

Correct Answer: 7

Suppose P(x₁, y₁) & Q(x₂, y₂)
Solve: 2x² – rx + p = 0 for x₁ and x₂
Solve: x² – sx – q = 0 for y₁ and y₂
So, the quation of circle = (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
x² – (x₁ + x₂)x + x₁x₂ + y² – (y₁ + y₂)y + y₁y₂ = 0
x2−\(\frac r2\)x+\(\frac p2\)+y²+sy−q = 0
2x² + 2y² – rx + 2sy + p – 2q = 0
On comparing with 2x² + 2y² – 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.