Question

If the area of the circle $4x^{2} + 4y^{2} - 8x + 16y + k = 0$ is $9\pi$ sq. units, then $k =$

• A. $4$
• B. $16$
• C. $ \pm16$
• D. $-16$

Answer 1

Answer: (D)

Equation of the circle is
$4x^{2} + 4y^{2} - 8x + 16y + k = 0$
$\Rightarrow \, x^{2} +y^{2} -2 x + 4y +\frac{k}{4}=0$
$\therefore $ Radius is $\sqrt{g^{2}+f ^{2}-c}=\sqrt{\left(-1\right)^{2}+\left(2\right)^{2}-\frac{k}{4}}=\sqrt{5-\frac{k}{4}}$
Area is $ \pi \left(\sqrt{5-\frac{k}{4}}\right)^{2} =9\pi$ (given)
$\Rightarrow \, \pi\left(5-\frac{k}{4}\right)=9\pi$
$\Rightarrow \, -4=\frac{k}{4}$
$\Rightarrow \, k=-16$.