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I found this problem on a thread on Stack overflow where it was posted as "job interview question". Unfortunately I cannot find the question. But I saved the picture and just cannot figure it out.

The problem is to calculate the area of the crescent.

enter image description here

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In almost any "diagram" problem about circles, putting a dot and a label at the center is helpful. We have a label for the center of the big circle. Put in the center of the smaller circle. –  André Nicolas Apr 29 '11 at 19:58
@user6312 I don't think the OP created this figure, as they indicate it was copied from another discussion. –  Matthew Conroy Apr 29 '11 at 20:17
Do you know what is amusing, I just had a candidate for an interview tell me they found the answer here. Because I posted the question on Programmers - didn't realise there was an answer supplied online! Oh well, research is the name of the game for my tests. –  Orbling Nov 6 '13 at 15:33

5 Answers 5

up vote 10 down vote accepted

Assuming AD is the diameter of the smaller circle and C is the center of the larger circle.

If $CD = x$ then, $CE = 4+x$.

Note that angle DEA is a right triangle.

We have by the similarity of triangles EDC and ACE that

$\frac{x}{4+x} = \frac{4+x}{9+x}$

Solving gives $x = 16$. Thus the radius of larger circle is $25$. The radius of the smaller circle is $\frac{x + 9+x}{2} = 20.5$

Area of the crescent = $\pi ((25)^2 - (20.5)^2) = 204.75 \times \pi$

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found the original post here:

and there is the solution in the comments:

the diameters differ by 9cm, so if the inner circle has radius r, the outer circle has radius r + 4.5. The area of the crescent is the difference in the areas of the circles: pi(r + 4.5)^2 - pi * r^2. All that's left is finding r. Define C as the point (0,0), then point E is at (0, r - 0.5) (because CE is 5cm less than the larger radius). The inner circle is shifted right 4.5cm, so its equation is (x - 4.5)^2 + y^2 = r^2. Plug in (x,y) = (0, r - 0.5) and solve for r.

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Here's how I'd do it:

Call $R$ the radius of the big circle and $r$ that of the small one. Now, observe that the surface of the crescent is just the difference between the surface of the big circle and the small one or

$$\pi R^2 - \pi r^2 = \pi (R-r)(R+r) \; .$$

Note how I've expressed this surface as a product of two quantities which I am now going to determine from the other data in the drawing. First the difference between the double radii is clearly 9cm:


We're halfway. Then the distance between the center of the small circle and point E is obviously $r$, but can alternatively be expressed with Pythagoras as

$$r^2 = (R-5)^2+(r-R+9)^2 \; .$$

Reordering and using what we already know about $R-r$:

$$r^2 - (R-5)^2= \left(\frac{9}{2}\right)^2 \; .$$

Again, using the factorizing trick

$$(r - R+5)(r+R-5)= \left(\frac{9}{2}\right)^2 \; .$$


$$r+R= 5+2\left(\frac{9}{2}\right)^2 \; .$$

Combining everything, we get that the surface of the crescent is

$$\pi \frac{9}{2}\left(5+2\left(\frac{9}{2}\right)^2\right)$$

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Here's my brute force, minimal geometry solution.

Let the smaller circle have center $(c,0)$ and radius $r$. Let $E$ have coordinates $(0,e)$ and $D$ have coordinates $(-d,0)$. Let $R$ be the radius of the larger circle.

Then we have five unknowns and the five equations $$ \begin{eqnarray} c^2+e^2 &=& r^2 \\ c+d &=& r \\ R &=& e+5 \\ R &=& d+9 \\ R &=& c+r \end{eqnarray} $$ which can be easily solved. One finds $c=4.5$, $d=16$, $e=20$, $r=20.5$, and $R=25$.

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For any chord passing through a point inside a circle, the product of the lengths of the sub-segments is a number depending only on the point, not the chord. (This number is called the "power" of that point, though the value itself isn't relevant here.) That is, if any two chords contain the same point, then the products of their respective sub-segments are equal.

In your diagram ---writing $E^\prime$ for the reflection of point $E$ in the (presumed) diameter $AD$--- we have chords $AD$ and $EE^\prime$ meeting at point $P$ inside the smaller circle. Therefore,

$$CA \cdot CD = CE \cdot CE^\prime \;\;\;\; \left(\;= \text{power of point}\;\; C\;\right)$$

Now, let $R$ be the radius of the big circle. Then $CA = R$, $CD = R-9$, and $CE=CE^\prime=R-5$. By the above,


so that $R = 25$. From there, it's easy.

The Power of a Point notion --which also relates not only chords, but tangents and secants that meet a point-- is a handy one to know.

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