If the area of the circle $4x^{2} + 4y^{2} - 8x + 16y + k = 0$ is $9\pi$ sq unit, then the4 value of $k$ is

Question

Dr. Abubacker · Accepted Answer

(c)
Given equation of circle can be rewritten as
$x^{2} + y^{2} - 2x + 4y + \frac{k}{4} = 0$
$	herefore Radius\ of\ circle = \sqrt{1 + 4 - \frac{k}{4}} = \sqrt{5 - \frac{k}{4}}$
$Area\ of\ circle = 9\pi$ (given)
$\Rightarrow \ \pi\left( 5 - \frac{k}{4} ight) = 9\pi$
$\Rightarrow \ \ 5 - 9 = \frac{k}{4} \Rightarrow k = - 16$

Q. If the area of the circle $4x^{2} + 4y^{2} - 8x + 16y + k = 0$ is $9\pi$ sq unit, then the4 value of $k$ is

Explanation

Reviewed by Dr. Abubacker

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