# 1960 AHSME Problems/Problem 7

## Problem

Circle $I$ passes through the center of, and is tangent to, circle $II$. The area of circle $I$ is $4$ square inches. Then the area of circle $II$, in square inches, is: $\textbf{(A) }8\qquad \textbf{(B) }8\sqrt{2}\qquad \textbf{(C) }8\sqrt{\pi}\qquad \textbf{(D) }16\qquad \textbf{(E) }16\sqrt{2}$

## Solutions $[asy] draw(circle((0,0),50)); draw(circle((-25,0),25)); [/asy]$

### Solution 1

Since Circle $I$ is tangent to circle $II$ and touches the center of circle $II$, the diameter of circle $I$ is the radius of circle $II$.

That means circle $II$ is twice as big as circle $I$, so the area of circle $II$ is four times as big as circle $I$.

The area of circle $II$ is $16$ square inches, so the answer is $\boxed{\textbf{(D)}}$.

### Solution 2

Since Circle $I$ is tangent to circle $II$ and touches the center of circle $II$, the diameter of circle $I$ is the radius of circle $II$.

Applying the area formula $A = \pi r^2$, substitute $4$ for $A$ to solve for the radius of circle $1$. $$4 = \pi r^2$$ $$\frac{4}{\pi} = r^2$$ $$r = \frac{2}{\sqrt{\pi}}$$

That means the diameter of circle $I$ (or the radius of circle $II$) is $\frac{4}{\sqrt{\pi}}$. Apply the area formula again to find the area of circle $II$. $$A = \pi (\frac{2}{\sqrt{\pi}})^2 = \pi \cdot \frac{4}{\pi} = 16$$

The answer is $\boxed{\textbf{(D)}}$.