Question

If the length of the tangent from any point on the circle ${(x-3)}^2+{(y+2)}^2=5r^2$ to the circle ${(x-3)}^2+{(y+2)}^2=r^2$ is $16$ units, then the area between the two circles in sq unit is

• A. $32\pi$
• B. $4\pi$
• C. $8\pi$
• D. $256\pi$

Answer 1

The correct option is D $256\pi$

Let point $P(x_1,y_1)$ be any point on the circle, therefore it satisfy the circle
${(x_1-3)}^2+{(y_1+2)}^2=5r^2$ ....(i)
The length of the tangent drawn from point $P(x_1,y_1)$ to the circle ${(x-3)}^2+{(y+2)}^2=r^2$ is
$\sqrt{{(x_1-3)}^2+{(y_1+2)}^2-r^2}=\sqrt{5r^2-r^2}$ .....[from equation (i)]
$\Rightarrow 16=2r$
$\Rightarrow r=8$
Therefore, the area between two circles
$=5\pi r^2-\pi r^2$
$=4\pi r^2$
$=4\pi \times 8^2$
$=256\pi$ sq units