Q93 / 160 · TS EAMCET 2013 · Mathematics

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
    $-42$
  • Correct answer
    D
    $-62$

Solution

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 the circle. $$ x^2+y^2-6 x+4 y-12=0 $$ Therefore, Eq. (i) passes through the centre of second circle i.e., $(3,-2)$ $$ \begin{aligned} & \therefore & 10(3)-10(-2)+c+12 & =0 \\ \Rightarrow & & 30+20+c+12 & =0 \\ \Rightarrow & & c & =-62 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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