JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let the abscissae of the two points \(P\) and \(Q\) on a circle be the roots of \(x^{2}-4 x-6=0\) and the ordinates of \(P\) and \(Q\) be the roots of \(y ^{2}+2 y -7=\) 0.If \(PQ\) is a diameter of the circle \(x ^{2}+ y ^{2}+2 ax +\) \(2 by + c =0\), then the value of \((a+b-c)\) is.
- A \(12\)
- B \(13\)
- C \(14\)
- D \(16\)
Answer & Solution
Correct Answer
(A) \(12\)
Step-by-step Solution
Detailed explanation
Equation of circle diameter form \(\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0\) (where \(x_{1}, x_{2}\) are the roots of \(x^{2}-4 x-6=0\) and \(y_{1}\), \(y _{2}\) are the roots of \(y ^{2}+2 y -7=0\) ) \(x^{2}+y^{2}-4 x+2 y-13=0\)…
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