Q61 / 180 · COMEDK 2012 · Mathematics
If the area of the circle $7 x^{2}+7 y^{2}-7 x+14 y+k-0$ is $12 \pi$ sq units, then the value of $k$ is
- A$\frac{-43}{4}$
- Correct answerB$\frac{-301}{4}$
- C$-16$
- D$\pm 4$
Solution
We have,
$$
\begin{aligned}
7 x^{2}+7 y^{2}-7 x+14 y+k &=0 \\
x^{2}+y^{2}-x+2 y+\frac{k}{7} &=0
\end{aligned}
$$
Here, $\quad C \equiv\left(\frac{1}{2},-1\right)$
and $\quad r=\sqrt{\frac{1}{4}+1-\frac{k}{7}}$
$$
r=\sqrt{\frac{5}{4}-\frac{k}{7}}
$$
Given that, Area $=12 \pi \mathrm{sq}$. unit
$\Rightarrow \quad 12 \pi=\pi\left(\frac{5}{4}-\frac{k}{7}\right)$
$\Rightarrow \quad 12=\frac{5}{4}-\frac{k}{7} \Rightarrow 12=\frac{35-4 k}{28}$
$\Rightarrow \quad 336=35-4 k$
$\Rightarrow \quad 4 k=-301 \Rightarrow k=-\frac{301}{4}$
Hence, the correct answer is option: 'B'.