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The red, blue, and green circles have diameters 3, 4, and 5, respectively.

What is the radius of the black circle tangent to all three of these circles?

I just figured out the radius is exactly $dfrac7223$ but I don't know how to do the solution.

Let's find tangent circle by applying inversion transformation, or equivalently working in complex domain and applying reciprocal transformation $w=1/overlinez$.

Let (0,0), or the inversion centre, be the intersection of 3 circles. Other circle intersections are $(0,3)$, $(4,0)$ and $frac1225left(3,4right)$.

Inversion transformation transform a circle passing through centre into a line, and other circles into circles. Thus, given three circles are transformed into lines, intersecting at 3 points; while big (still unknown) tangent circle transforms into circle tangent on those three lines. Transformed intersection coordinates are:

$ beginarraylcr (0,3) && longrightarrow&& left(0,frac13right)\ (4,0) && longrightarrow && left(frac14,0right)\ frac1225left(3,4right) && longrightarrow && left(frac14,frac13right)endarray $

There are 4 circles that are tangent on all 3 lines: 1 inscribed and 3 escribed circle. The correct one has (0,0) in its interior. Finding coordinates of the centre and radius (solving quadratic equation) gives following circle:

$c=left(-frac16,-frac14right)quad r=frac12$

Transforming back escribed circle gives the required tangent circle. Take notice that, while points on a circle transform into a circle, a circle centre does not transform to a corresponding circle centre. Radius of transformed circle can be deduce by working on the line that connects origin and centre of the circle. Transforming 2 points of the circle that are collinear with circle centre and the origin gives.

$2R= left|frac1c-frac1cright|\ R= fracrleft\ R=frac7223$

Here is the picture with requested tangent circle, as well as one other tangent circle corresponding to the inscribed circle of the reciprocal space.

Radius is exactly $dfrac7223$, pretty neat. See the figure below, where dotted lines from the center of the circumcribing circle passes through the midpoints of the sides of the triangle.

Let : $B$ be the origin $(0,0)$, and the center of circle of unknown radius $r$ be $(x,y)$. Then we solve the following three equations to find $r$

$dfrac32+sqrtx^2+(dfrac32-y)^2=r$

$2+sqrt(2-x)^2+y^2=r$

$dfrac52+sqrt(2-x)^2+(dfrac32-y)^2=r$

so that $(r, x, y)= left(dfrac7223, dfrac3623, dfrac2423right)$

Note: The midpoints of the sides $(0,dfrac32)$, $(2,0)$, and $(2,dfrac32)$ are centers of circles of radii $dfrac32, 2, dfrac52$.
Given any two tangent circles, their centres and point of tangency are colinear.

Bonus: If right triangle : $triangle ABC$ has rational sides, then $(r,x,y)$ are also all rationals.

Thank you for your answers that helped me make a solution

$M_red=dbinom03/2$ , $r_red=dfrac32$ , $M_blue=dbinom02$ , $r_blue=2$, $M_green=dbinom23/2$ , $r_green=dfrac52$

$M_black=M=dbinomxy$ , $r_black=r$

$length(v)=dfracvsqrtv_x^2+v_y^2$ , $unitvec(v)=dfracvlength(v)$

$d_red=length[M_red+unitvec(M_red-M)times r_red-M]$

$d_blue=length[M_blue+unitvec(M_blue-M)times r_blue-M]$

$d_green=length[M_green+unitvec(M_green-M)times r_green-M]$

$dbinom03/2+dfracdbinom03/2-dbinomxysqrt(0-x)^2+(3/2-y)^2times dfrac32-M$ $Rightarrow $ $d_red=sqrtleft(-x-dfrac3xsqrt4x^2+(3-2y)^2right)^2+left(3/2-y+dfrac9-6y2sqrt4x^2+(3-2y)^2right)^2$

$dbinom20+dfracdbinom20-dbinomxysqrt(2-x)^2+(0-y)^2times 2-M$ $Rightarrow $ $d_blue=sqrtleft(2-x+dfrac4-2xsqrt(-2+x)^2+y^2right)^2+left(-y-dfrac2ysqrt(-2+x)^2+y^2right)^2$

$dbinom23/2+dfracdbinom23/2-dbinomxysqrt(2-x)^2+(3/2-y)^2times dfrac52-M$ $Rightarrow $ $d_red=sqrtleft(2-x-dfrac5(-2+x)sqrt25+4x(-4+x)+4y(-3+y)right)^2+left(3/2-y+dfrac5(3/2-y)2sqrt(-2+x)^2+(-3/2+y)^2right)^2$

Setting these 3 equations to r and solving for x,y,r of the black circle results in:

$x=dfrac3623 , y=dfrac2423 , r=dfrac7223$

It is simply the Apollonius Problem and solved by the following three quadratic equations using WolframAlpha for center $ (x,y) $ and radius $r$ of the biggest circle, which is tangent to other three circles;

(1) $ x^2 + (y - frac32)^2 = (r - frac32)^2 $, circle on side of length $ = 3 $,

(2) $ (x - 2)^2 + y ^ 2 = (r - 2)^2 $, circle on side of length $ = 4 $,

(3) $ (x - 2)^2 + (y - frac32)^2 = (r - frac52)^2 $, circle on side of length $ = 5 $.

The solution is $ x = frac3623, y = frac2423 $ and radius $ r $ is the Answer $ = frac7223 $.