Q82 / 180 · COMEDK 2020 · Mathematics
If the circles $x^{2}+y^{2}-2 x-2 y-7=0$ and $x^{2}+y^{2}+4 x+2 y+k=0$ cut orthogonally, then the length of the common chord of the circles is
- A2
- Correct answerB$12 / \sqrt{13}$
- C8
- D5
Solution
We have,
$$
\begin{aligned}
&S_{1}: x^{2}+y^{2}-2 x-2 y-7=0 \\
&S_{2}: x^{2}+y^{2}+4 x+2 y+k=0
\end{aligned}
$$
Since, $S_{1}$ and $S_{2}$ cut orthogonally, then
$$
\begin{aligned}
& & 2 g_{1} E_{2}+2 f_{1} f_{2} &=c_{1}+c_{2} \\
\Rightarrow & 2(-1)(2)+2(-1)(1) &=-7+k \\
\Rightarrow &-4-2 &=-7+k \\
k^{\prime} &=1
\end{aligned}
$$
Now, let $r_{1}$ and $r_{2}$ be the radius of $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$, respectively. Then,
$$
\begin{aligned}
&n_{1}=\sqrt{(-1)^{2}+(-1)^{2}-(-7)}=3 \\
&r_{2}=\sqrt{(2)^{2}+(1)^{2}-1}=2
\end{aligned}
$$
Now, length of common chord
$$
\begin{aligned}
&=\frac{2 r_{2}}{\sqrt{r_{1}^{2}+r_{2}^{2}}} \\
&=\frac{2 \times 3 \times 2}{\sqrt{(3)^{2}+(2)^{2}}}=\frac{12}{\sqrt{13}}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.