JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let the abscissae of the two points \(P\) and \(Q\) be the roots of \(2 x ^{2}- rx + p =0\) and the ordinates of \(P\) and \(Q\) be the roots of \(x^{2}-s x-q=0\). If the equation of the circle described on \(PQ\) as diameter is \(2\left( x ^{2}+ y ^{2}\right)-11 x -14 y -22=0\), then \(2 r+s-2 q+p\) is equal to
- A \(5\)
- B \(9\)
- C \(6\)
- D \(7\)
Answer & Solution
Correct Answer
(D) \(7\)
Step-by-step Solution
Detailed explanation
Equation of the circle with \(PQ\) as diameter is \(2\left(x^{2}+y^{2}\right)-r x-2 s y+p-2 q=0\) on comparing with the given equation \(r\) = \(11\), \(s\) = \(7\) \(p-2 q=-22\) \(\therefore 2 r+s-2 q+p=22+7-22=7\)
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