AP EAMCET · Maths · Circle
The equation of the circle which passes through the point \((3,2)\) bisects the circumference of the circle \(x^2+y^2=15\) and cuts the circle \(x^2+y^2+4 x+6 y+3=0\) orthogonally is
- A \(x^2+y^2+6 x+8 y-43=0\)
- B \(x^2+y^2+6 x-8 y-15=0\)
- C \(x^2+y^2-6 x+8 y-11=0\)
- D \(x^2+y^2-6 x-8 y+21=0\)
Answer & Solution
Correct Answer
(B) \(x^2+y^2+6 x-8 y-15=0\)
Step-by-step Solution
Detailed explanation
Let the equation of required circle is \[ x^2+y^2+2 g x+2 f y+c=0 \] Since, circle (i) passes through point \((3,2)\), so \[ \begin{aligned} & & 9+4+6 g+4 f+c & =0 \\ \Rightarrow & & 6 g+4 f+c+13 & =0 \end{aligned} \] Since, circle (i) bisects the circumference of the circle…
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