# Radius of the circle internally tangent to three circles whose diameters are the sides of a $3$-$4$-$5$ right triangle

Given a right triangle with side lengths $$3$$, $$4$$, and $$5$$, a circle is drawn with each side as a diameter. Find the radius of the circle that is internally tangent to all three circles.

Would radical axes be a valid approach here? I am thinking coordinates at the moment, as well. I haven't really made any progress with this problem as it's very hard.

Any solutions not involving advanced geometry techniques?

UPDATE: Graphing isn't working either. I can't see anything that might be useful. • Please read tag descriptions before applying them: the algebraic geometry tag expressly states that "[t]his tag should not be used for elementary problems which involve both algebra and geometry." – KReiser Aug 26 at 23:41
• I'm sorry. I will do so in the future; I'm quite new here. – Fleccerd Aug 26 at 23:49
• This is the Problem of Apollonius, the tenth case in the table. – Ross Millikan Aug 27 at 1:46

I think the best naive approach is to place the figure on a coordinate plane. To make the numbers nice, it might be better to scale up the triangle by a factor of $$2$$, then when we find the circle, scale everything back down.

The most natural way to proceed, I think, is to let the midpoint of the hypotenuse be the origin, and the right angle is at $$C = (4,3)$$. Then the centers of the semicircles are at $$C' = (0,0)$$, $$B' = (4,0)$$, and $$A' = (0,3)$$. We are looking for a point $$P = (x,y)$$ such that $$PC' + 5 = PB' + 3 = PA' + 4 = r$$, where $$r$$ is the radius of the tangent circle. This leads to the system of equations \begin{align} (r-5)^2 &= x^2 + y^2 \\ (r-3)^2 &= (x-4)^2 + y^2 \\ (r-4)^2 &= x^2 + (y-3)^2. \end{align} Consequently $$4(4-r) = (r-5)^2 - (r-3)^2 = x^2 - (x-4)^2 = 8(x-2),$$ or $$x = \frac{8-r}{2}$$, and similarly $$9-2r = (r-5)^2 - (r-4)^2 = y^2 - (y-3)^2 = 3(2y-3),$$ or $$y = \frac{9-r}{3}$$. Thus $$(r-5)^2 = \left(\frac{8-r}{2}\right)^2 + \left(\frac{9-r}{3}\right)^2,$$ for which the unique positive root is $$r = \frac{144}{23}$$, hence after undoing the scaling, the desired radius is $$\boxed{r = \frac{72}{23}}.$$ The center is located at $$P = (\frac{10}{23}, \frac{21}{46})$$.

Whether we can find a more elegant solution, or through purely geometric means, remains an open question, but this approach I find quite straightforward, elementary, and not computationally difficult.

Since two users have disputed my answer, I show the following Mathematica code and Geogebra figure:

First[{x, y, r}/2 /. Solve[{EuclideanDistance[{x, y}, {0, 0}] + 5 ==
EuclideanDistance[{x, y}, {4, 0}] + 3 ==
EuclideanDistance[{x, y}, {0, 3}] + 4 == r}, {x, y, r}]]


The output is

{10/23, 21/46, 72/23}


The figure is shown below. • I think there's some mistake. When I tried to sketch this in Geogebra, I'm not getting the desired result. – SarGe Aug 27 at 13:52
• With @heropup's method, I'm getting $r = 72/23$ and center $P = (36/23, 24/23)$ – cosmo5 Aug 27 at 15:30
• @cosmo5 You have ignored the fact that I have rotated the figure. – heropup Aug 27 at 16:59 Let the center of the circle be $$(x, y)$$ and radius of the circle be $$r$$.

Since, the circle touches all the three circles, we get the constraints as \begin{align}(x-2)^2+y^2&=(2-r)^2\\ x^2+(y-1.5)^2&=(1.5-r)^2\\ (x-2)^2+(y-1.5)^2&=(2.5-r)^2 \end{align}

Solving gives us $$\displaystyle (x,y,r)=\left(\frac{36}{23},\frac{24}{23},\frac{72}{23}\right),(0,0,0)$$.