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. units is

• A. $16\pi$
• B. $8\pi$
• C. $4\pi$
• D. $256\pi$

Answer 1

Answer: (D)

Let point $P(x_1,y_1)$ be any point on the circle, therefore it satisfy the circle
$(x_1-3{)}^2+(y_1+2{)}^2=5r^2\dots (i)$
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
$\sqrt{{\left(x_1-3\right)}^2+{\left(y_1+2\right)}^2-r^2}$
$=\sqrt{5r^2-r^2}$ (From (i))
$\Rightarrow 16=2r$
$\Rightarrow r=8$
$\therefore$ The area between two circles

$=\pi 5r^2-\pi r^2$
$=4\pi r^2=4\pi \times 8^2$
$=256\pi$ sq unit