Found here (you can peek for a solution there), source is (again) "Geometria Super Top" on Facebook.
What is the diameter of the circle in the following diagram? All marked angles are $90^\circ$.
Personally, I find the solution given there too complicated; I did it in a minute with nothing but good old Pythagoras. (Verified with MATHEMATICA)
Bonus: What happens if we add a fourth perpendicular line? (Generalize to a,b,c,d as lengths.)
Using nothing but good old Pythagoras:
Construct two right triangles with side lengths x, 14, r and 15 - x, 6, r.
By Pythagoras,
r2 = x2 + 142 = (15 - x)2 + 62
⇒ x = (152 + 62 - 142) ÷ (2 · 15) = 13/6.
The first right triangle is a 13–84–85 triangle scaled down by 6, so r = 85/6 and the diameter is 85/3.
A straightforward solution.
We have to find the diameter of the circle passing through three given points.
That is, the diameter $D$ of the circle which is circumscribed at the triangle whose vertices are the given points.
By Pythagoras' theorem, the lengths $x$, $y$, $z$ of the sides of the triangle are $12$, $\sqrt{15^2+8^2}=17$, and $\sqrt{15^2+20^2}=25$, respectively.
The semiperimeter $s=\frac{x+y+z}2$ of the triangle equals $27$. By Heron's formula, the triangle has area $A=\sqrt{s(s-x)(s-y)(s-z)}=\sqrt{27\cdot 15\cdot 10\cdot 2}=90$.
Finally, the diameter $D$ equals $\frac{xyz}{2A}=\frac{12\cdot 17\cdot 25}{180}=\frac{85}{3}$.
Bonus.
In this case the lengths $x$, $y$, $z$ of the sides of the triangle are $\sqrt{a^2+b^2}$, $\sqrt{c^2+d^2}$, and $\sqrt{(a+c)^2+(b+d)^2}$, respectively. Let $\gamma$ be the angle between the first two sides. Then the diameter $D$ of the circumscribed circle equals $\frac{z}{\sin\gamma}$. But $z^2=x^2+y^2-2xy\cos \gamma $, so $$(a+c)^2+(b+d)^2= a^2+b^2+ c^2+d^2-2\sqrt{(a^2+b^2)(c^2+d^2)}\cos\gamma,$$ $$\cos\gamma =-\frac{ ac+bd}{\sqrt{(a^2+b^2)(c^2+d^2)}},$$ $$\sin\gamma=\sqrt{1-\cos^2\gamma}=\frac{|ad-bc|}{\sqrt{(a^2+b^2)(c^2+d^2)}}.$$ Finally, $$D=\frac{\sqrt{(a^2+b^2)(c^2+d^2)( (a+c)^2+(b+d)^2)}}{|ad-bc|}.$$
Call the point where the red line meets the circle $A$, the point where the blue and yellow lines meet $B$, and the other point where the yellow line meets the circle $C$. Call the intersection of the red and blue lines $O$.
Then:
Extend $AO$ to meet the circle at $D$. By symmetry about a horizontal axis through the midpoint of $BC$, we see that $OD$ has length 8 + 12 = 20. Extend $BO$ to meet the circle at $E$. By the intersecting chords theorem, we know $BO \cdot OE = AO \cdot OD$. We calculate $OE = 32/3$. Notice that $CE$ is a diameter since it subtends a right angle at $B$. Calculate the length of $CE$ using Pythagoras. The answer is $85/3$
Consider triangle $\triangle ABC$. Its $A$-altitude is of length $y$, and hence its area is $$[\triangle ABC]=\frac{yz}{2}.$$
Let $a=BC$, $b=CA$, and $c=AB$ as usual. Then $a=z$, $b=\sqrt{(x+z)^2+y^2}$, and $c=\sqrt{x^2+y^2}$ by Pythagoras. Let $R$ be the radius of the circle (i.e. the circumcircle of $\triangle ABC$). It is well known (say, by the Law of Sines) that $[\triangle ABC]=\frac{abc}{4R}$, from which plugging in we get $$\frac{yz}2=\frac{z\sqrt{(x^2+y^2)((x+z)^2+y^2)}}{4R}\implies \boxed{R=\frac{\sqrt{(x^2+y^2)((x+z)^2+y^2)}}{2y}}.$$
Plug in $(x,y,z)=(8,15,12)$, we have $$R=\frac{\sqrt{(x^2+y^2)((x+z)^2+y^2)}}{2y}=\frac{\sqrt{(8^2+15^2)(20^2+15^2)}}{2\cdot 15}=\frac{17\cdot25}{30}=\frac{85}6.$$
