# Three Circles One Area

We have three circles where their centers are on the same line and PR is tangent to both small circles as shown below.

If $|PR|=12$ unit, what is the area of blue part of the circle?

Reference: A Turkish Journal: Bilim Teknik

• Are the circles in a ratio of size to each other? Nov 13 '17 at 17:23
• as in is the smallest circle half the size of the second largest etc... Nov 13 '17 at 17:27
• No, I didn't. Perhaps because you made your comment and then deleted it before I visited PSE again. But, seriously, why delete it? (Unless you were trying to avoid letting anyone see that you got the puzzle from somewhere else, but that would be pretty silly.) Nov 14 '17 at 11:42
• And I hope you won't stop posting questions. But you do need to give appropriate credit for any questions you take from elsewhere! Nov 14 '17 at 12:08
• @AlwaysConfused Ah, but it does tell you something that constrains the size of the white circles, namely the length of that tangent line. And, as it happens, that (plus the fact that the centres are collinear) is enough to determine the answer, even though the size of each individual circle can vary. Sep 11 '19 at 15:12

If the question has a well defined answer then

it must not depend on which of the infinitely many configurations satisfying the given description we choose.

So

we can take PR to be a diameter of the large circle, which then has area $36pi$; $2\times\frac14=\frac12$ of the circle is white, hence $\frac12$ is blue, so the blue area is $18\pi$.

That's a bit cheeky, of course. So here's a more conventional solution.

Let Q be the point of tangency in the middle of that common tangent. Let A,B be the two ends of the diameter passing through Q. Then we have $AP^2=PQ^2+d_1^2$ and $BP^2=PQ^2+d_2^2$ where $d_1,d_2$ are the diameters of the two white circles. And now note that since the angle in a semicircle is always $\pi/2$ we have $AP^2+BP^2=d^2$ where $d$ is the diameter of the outer circle. Putting this together, $d^2=2PQ^2+d1^2+d2^2$; our area is $\frac\pi4(d^2-d_1^2-d_2^2)=\frac\pi2\cdot PQ^2=\frac\pi8PR^2$ and we're done.

• Smart but cheaty. I added a slightly less smart but less cheaty solution. Nov 13 '17 at 17:40
• A problem with a similar feel to it and a similar "smart" solution: A sphere has a hole bored along a diameter. The length of the hole (measured beteween its actual ends, not the now-removed poles of the sphere at the ends of that diameter) is, say, 12 units. What is the volume of what remains? Cheaty answer: Suppose the hole is of zero thickness; then 12 units is the diameter of the sphere and no material has been removed. Etc. Nov 13 '17 at 17:42
• @GarethMcCaughan Is it not correct to say that PR is diameter of the large circle, whose centre(Q, as said by you or the middle dot in the picture) is not on PR? Nov 14 '17 at 7:08
• In the diagram here, unless the small circles are equal PR is not a diameter. But the perpendicular to it through Q is a diameter. Nov 14 '17 at 10:58

$18\pi$

Let $d_s=$ diameter of small inner circle and $d_l=$ diameter of large inner circle,
then $d$ (diameter of outer circle) $=$

$d_s+d_l$
and, of course, $r=\frac{d_s+d_l}{2}$

Using $d$ and the length of PR, we can find the value of $d_s$ and $d_l$ as following :

PR is a chord and its equation is :
Chord length $=2\sqrt{r^2-l^2}$, where $r$ is the radius of a circle and $l$ is the perpendicular distance between a chord and the center of a circle.
After inspecting the circle, we can find that $l=d_l-r$, diameter of larger inner circle - radius of outer circle.
So,
PR$=12=2\sqrt{r^2-(d_l-r)^2}$
$=>6=\sqrt{(\frac{d_s+d_l}{2})^2-(\frac{2d_l}{2}-\frac{d_s+d_l}{2})^2}$
$=\sqrt{(\frac{d_s+d_l}{2})^2-(\frac{d_l-d_s}{2})^2}$
$=\sqrt{\frac{d_s^2+2d_sd_l+d_l^2}{4}-\frac{d_l^2-2d_sd_l+d_s^2}{4}}$
$=\sqrt{\frac{4d_sd_l}{4}}$
$=\sqrt{d_sd_l}$
$=>36=d_sd_l$

Since we know the value of $d_s$ and $d_l$ now, we are ready to calculate the area of shaded area :

Area of a circle $=\pi r^2$.
Using this equation, we can find that
Area of outer circle $=\pi(\frac{d_s+d_l}{2})^2=\frac{\pi}{4}(d_s+d_l)^2$
Area of small inner circle $=\pi(\frac{d_s}{2})^2=\frac{\pi}{4}d_s^2$
Area of large inner circle $=\pi(\frac{d_l}{2})^2=\frac{\pi}{4}d_l^2$
and
Area of shaded area $=$ Area of outer circle $-$ Area of small inner circle $-$ Area of large inner circle
$=\frac{\pi}{4}[(d_s+d_l)^2-d_s^2-d_l^2]$
$=\frac{\pi}{4}[(d_s^2+2d_sd_l+d_l^2)-d_s^2-d_l^2]$
$=\frac{\pi}{4}[2d_sd_l]$
and we already know what $d_sd_l$ is from above :)
Therefore, Area of shaded area $=\frac{\pi}{4}[2*36]=18\pi$