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
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Sign up to join this communityWe 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
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.
Answer is
$18\pi$
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 $=\frac{\pi}{4}[2*36]=18\pi$