TS EAMCET · Maths · Circle
If the circle \(x^2+y^2+4 x-6 y+c=0\) bisects the circumference of the circle \(x^2+y^2-6 x+4 y-12=0\), then \(c\) is equal to
- A \(16\)
- B \(24\)
- C \(-42\)
- D \(-62\)
Answer & Solution
Correct Answer
(D) \(-62\)
Step-by-step Solution
Detailed explanation
The common chord of the given circle is \[ \begin{array}{cc} & S_1-S_2=0 \\ \Rightarrow & \left(x^2+y^2+4 x-6 y+c\right) \\ & -\left(x^2+y^2-6 x+4 y-12\right)=0 \\ \Rightarrow & 10 x-10 y+c+12=0 \end{array} \] Since, circle \(x^2+y^2+4 x+6 y+c=0\) bisects the circumference of…
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