Q93 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
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$x^2+y^2+8 x+6 y-12=0$
- Correct answerC$x^2+y^2+6 x+8 y-12=0$
- D$x^2+y^2-6 x-8 y-12=0$
Solution
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 $\left(x^2+y^2+4 x+6 y-12\right)$
$$
\begin{array}{r}
+\lambda\left(x^2+y^2-6 x-4 y-12\right)=0 \\
\Rightarrow x^2+y^2+\frac{4-6 \lambda}{1+\lambda} x+\frac{6-4 \lambda}{1+\lambda}-12=0 \quad \ldots \text { (iii) }
\end{array}
$$
Since, the another circle
$$
x^2+y^2-4 x+2 y+8=0
$$
cuts the circle (iii) orthogonally, then
$$
\begin{aligned}
& -2\left(\frac{4-6 \lambda}{1+\lambda}\right)+\left(\frac{6-4 \lambda}{1+\lambda}\right)=8-12=-4 \\
\Rightarrow & -8+12 \lambda+6-4 \lambda=-4-4 \lambda \\
\Rightarrow & 12 \lambda=-2 \Rightarrow \lambda=-\frac{1}{6}
\end{aligned}
$$
So, required equation of circle is
$$
\begin{aligned}
& 6\left(x^2+y^2+4 x+6 y-12\right) \\
& -\left(x^2+y^2-6 x-4 y-12\right)=0 \\
& \Rightarrow \quad 5 x^2+5 y^2+30 x+40 y-60=0 \\
& \Rightarrow \quad x^2+y^2+6 x+8 y-12=0 \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'C'.