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In the figure above, the ratio of the areas of the two circles touchin
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07 Aug 2025, 09:20
07 Aug 2025, 09:20
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In the figure above, the ratio of the areas of the two circles touching each other as shown is $2: 1$. If the area of the bigger circle is $\pi$, what is the distance between the centers of the two circles?
(A) $\(\frac{1+\sqrt{2}}{2}\)$
(B) $\(\frac{2+\sqrt{2}}{2}\)$
(C) $\(\frac{3+\sqrt{2}}{2}\)$
(D) $\(\frac{2+2 \sqrt{2}}{2}\)$
(E) 2
Re: In the figure above, the ratio of the areas of the two circles touchin
[#permalink]
20 Aug 2025, 08:54
20 Aug 2025, 08:54
1
Expert Reply
Let the radius of the bigger circle and the smaller circle be $\(R \& r\)$ respectively.
We are given that the area of the bigger circle is $\(\pi=\pi R^2 \Rightarrow R=\sqrt{1}=1\)$ $\(\qquad\)$
Also it is known that the area of the bigger circle to the smaller circle is in ratio $\(2: 1\)$, so we get
$$
\(\pi R^2: \pi r^2=2: 1 \Rightarrow R: r=\sqrt{2}: 1\)
$$
Using (1) \& (2), we get $\(\frac{R}{r}=\frac{\sqrt{2}}{1}=\frac{1}{r} \Rightarrow r=\frac{1}{\sqrt{2}}\)$
So, the distance between the centers of the two circles is sum of their radii
$$
\(=\mathrm{R}+\mathrm{r}=1+\frac{1}{\sqrt{2}}=\frac{\sqrt{2}+1}{\sqrt{2}}\)
$$
Hence the answer is (B).
We are given that the area of the bigger circle is $\(\pi=\pi R^2 \Rightarrow R=\sqrt{1}=1\)$ $\(\qquad\)$
Also it is known that the area of the bigger circle to the smaller circle is in ratio $\(2: 1\)$, so we get
$$
\(\pi R^2: \pi r^2=2: 1 \Rightarrow R: r=\sqrt{2}: 1\)
$$
Using (1) \& (2), we get $\(\frac{R}{r}=\frac{\sqrt{2}}{1}=\frac{1}{r} \Rightarrow r=\frac{1}{\sqrt{2}}\)$
So, the distance between the centers of the two circles is sum of their radii
$$
\(=\mathrm{R}+\mathrm{r}=1+\frac{1}{\sqrt{2}}=\frac{\sqrt{2}+1}{\sqrt{2}}\)
$$
Hence the answer is (B).
gmatclubot
Re: In the figure above, the ratio of the areas of the two circles touchin [#permalink]
20 Aug 2025, 08:54
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