AP EAMCET · Maths · Circle
The equation of the circle passing through the points of intersection of the circles \(x^2+y^2+4 x+6 y-12=0\) and \(x^2+y^2-6 x-4 y-12=0\) and cutting the circle \(x^2+y^2-4 x+4 y+8=0\) orthogonally is
- A \(x^2+y^2+6 x+8 y+12=0\)
- B \(x^2+y^2+8 x+6 y-12=0\)
- C \(x^2+y^2+6 x+8 y-12=0\)
- D \(x^2+y^2-6 x-8 y-12=0\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2+6 x+8 y-12=0\)
Step-by-step Solution
Detailed explanation
Equation of the circle passing through the points of intersection of the circles \[ \text { and } \quad \begin{aligned} x^2+y^2+4 x+6 y-12 & =0 \\ x^2+y^2-6 x-4 y-12 & =0 \end{aligned} \] and is \(\left(x^2+y^2+4 x+6 y-12\right)\)…
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