Question

If the area of the circle $7x^2+7y^2-7x+14y+k=0$ is $12\pi sq$. units, then the value of k is

• A. $\frac{-43}{4}$
• B. $\frac{-301}{4}$
• C. $-16$
• D. $\pm 4$

Answer 1

Answer: (B)

Equation of circle
$7x^2+7y^2-7x+14y+k=0$
can be written as.
$x^2+y^2-x+2y+\frac{k}{7}=0$
$\therefore$ Centre of circle is $\left(\frac{1}{2},-1\right)$
Radius of circle, $r=\sqrt{\frac{1}{4}+1-\frac{k}{7}}$
$=\sqrt{\frac{5}{4}-\frac{k}{7}}\Rightarrow r^2=\frac{5}{4}-\frac{k}{7}$
Given that area of circle = 12$\pi$ sq. units
$\Rightarrow \pi r^2=\pi \left(\frac{5}{4}-\frac{k}{7}\right)=12\pi$
$\Rightarrow \frac{5}{4}-\frac{k}{7}=12$
$\Rightarrow \frac{5}{4}-12=\frac{k}{7}\Rightarrow \frac{-43}{4}=\frac{k}{7}$
$\Rightarrow k=\frac{-301}{4}$