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

The correct option is D $-16$

Given equation of circle is $4x^2+4y^2-8x+16y+k=0⟹x^2+y^2-2x+4y+\frac{k}{4}=0$

Radius of the circle is $r=\sqrt{(-1{)}^2+(2{)}^2-\frac{k}{4}}=\sqrt{\frac{20-k}{4}}$

Given Area of the circle is $9\pi sq.units$

$⟹\pi r^2=9\pi ⟹\frac{20-k}{4}=9⟹20-k=36⟹k=-16$