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. $\pm 16$
• D. $-16$

Answer 1

Answer: (D)

Equation of the circle is
$4x^2+4y^2-8x+16y+k=0$
$\Rightarrow x^2+y^2-2x+4y+\frac{k}{4}=0$
$\therefore$ 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)
$\Rightarrow \pi \left(5-\frac{k}{4}\right)=9\pi$
$\Rightarrow -4=\frac{k}{4}$
$\Rightarrow k=-16$.