JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
The circle passing through the intersection of the circles, \(x^{2}+y^{2}-6 x=0\) and \(x^{2}+y^{2}-4 y=0\) having its centre on the line, \(2 x-3 y+12=0\), also passes through the point
- A \((1,-3)\)
- B \((-1,3)\)
- C \((-3,1)\)
- D \((-3,6)\)
Answer & Solution
Correct Answer
(D) \((-3,6)\)
Step-by-step Solution
Detailed explanation
Let \(S\) be the circle pasing through point of intersection of \(S _{1} \& S _{2}\) \(\therefore S = S _{1}+\lambda S _{2}=0\) \(\Rightarrow S :\left( x ^{2}+ y ^{2}-6 x \right)+\lambda\left( x ^{2}+ y ^{2}-4 y \right)=0\)…
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