Q88 / 180 · COMEDK 2016 · Mathematics
The equation of the common tangent of the two touching circles, $y^{2}+x^{2}-6 x-12 y+37=0$ and $x^{2}+y^{2}-6 y+7=0$ is
- A$x+y+5=0$
- B$x+y-5=0$
- C$x-y+5=0$
- Correct answerD$x-y-5=0$
Solution
Let
$\begin{aligned}
S_{1} & \equiv x^{2}+y^{2}-6 x-12 y+37=0 \\
\text { and } S_{2} & \equiv x^{2}+y^{2}-6 y+7=0
\end{aligned}$
The equation of common tangent of the two circles is
$\begin{aligned}
&\mathrm{S}_{1}-\mathrm{S}_{2}=0 \\
&\Rightarrow x^{2}+y^{2}-6 x-12 y+37-\left(x^{2}+y^{2}-6 y+7\right)=0 \\
&\Rightarrow-6 x+6 y+30=0 \\
&\Rightarrow x-y-5=0
\end{aligned}$
Hence, the correct answer is option: 'D'.