AP EAMCET · Maths · Circle
If the circle \(x^2+y^2+6 x-2 y+k=0\) bisects the circumference of the circle \(x^2+y^2+2 x-6 y-15=0\), then ' \(k\) ' is equal to
- A 21
- B –21
- C 23
- D –23
Answer & Solution
Correct Answer
(D) –23
Step-by-step Solution
Detailed explanation
Given, equation of circles are \[ \begin{aligned} & S: x^2+y^2+6 x-2 y+k=0 \\ & S^{\prime}: x^2+y^2+2 x-6 y-15=0 \end{aligned} \] Centre \(=(-1,3)\) Equation of common chaod is \(S-S^{\prime}=0\) \[ 4 x+4 y+(k+15)=0 \] Centre of \(S^{\prime}\) lies on the above line…
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