Question

The length of the tangent from any point on the circle $(x-3{)}^2+(y+2{)}^2=5r^2$ to the circle $(x-3)+(y+2{)}^2=r^2$ is $4$ units. Then the area between the circles is

• A. $32\pi$
• B. $4\pi$
• C. $8\pi$
• D. $16\pi$

Answer 1

Answer: (D)

$16\pi$

Let,p $\left[\right(3+\sqrt{5}r\cos \theta ),(-2+\sqrt{s}r\sin \theta \left)\right]$ be any point on
$(x-3{)}^2+(y+2{)}^2=5r^2$
then length of tangent from p on circle $(x-3{)}^2+(y+2{)}^2-r^2$ is $4$
$\therefore L=4=\sqrt{(\sqrt{5}r\cos \theta {)}^2+(\sqrt{5}r\sin \theta {)}^2-r^2}$
$=\sqrt{5r^2-r^2}=2r$
$r=2$
Given two circles are concentric with centre at $(3,-2)$
So, area between circles $O=\pi R^2-\pi r^2$
$\pi \left(\right(\sqrt{5}r{)}^2-r^2)$
$\pi 4r^2$
$\pi \times 4\times {}^{\circ }4$
$=16\pi$
correct answer is D.