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(D) Let point $P(x_1, y_1)$ be any point on the circle $(x-3)^2+(y+2)^2=5r^2$. Since $P$ lies on this circle, it satisfies the equation: $(x_1-3)^2+(y

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$256$

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(D) Let point $P(x_1, y_1)$ be any point on the circle $(x-3)^2+(y+2)^2=5r^2$. Since $P$ lies on this circle, it satisfies the equation: $(x_1-3)^2+(y_1+2)^2=5r^2 \dots (i)$ 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 given by $\sqrt{S_1}$, where $S_1 = (x_1-3)^2+(y_1+2)^2-r^2$. Substituting $(i)$ into the expression: Length $= \sqrt{5r^2-r^2} = \sqrt{4r^2} = 2r$. Given that the length of the tangent is $16$ units, we have $2r = 16$, which implies $r = 8$. The area between the two concentric circles is the difference of their areas: Area $= \pi(R^2) - \pi(r^2) = \pi(5r^2) - \pi(r^2) = 4\pi r^2$. Substituting $r = 8$: Area $= 4 \pi (8)^2 = 4 \pi (64) = 256 \pi$ sq. units.

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 (in $\pi$)

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