TITLE: What is the radius of the black circle tangent to all three of these circles?
QUESTION [28 upvotes]: 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 $\dfrac{72}{23}$ but I don't know how to do the solution.

REPLY [6 votes]: Thank you for your answers that helped me make a solution
$M_{red}=\dbinom{0}{3/2}$ , $r_{red}=\dfrac{3}{2}$ , $M_{blue}=\dbinom{0}{2}$ , $r_{blue}=2$, $M_{green}=\dbinom{2}{3/2}$ , $r_{green}=\dfrac{5}{2}$
$M_{black}=M=\dbinom{x}{y}$ , $r_{black}=r$
$length(v)=\dfrac{v}{\sqrt{v_{x}^{2}+v_{y}^{2}}}$ , $unitvec(v)=\dfrac{v}{length(v)}$
$d_{red}=length[M_{red}+unitvec(M_{red}-M)\times r_{red}-M]$
$d_{blue}=length[M_{blue}+unitvec(M_{blue}-M)\times r_{blue}-M]$
$d_{green}=length[M_{green}+unitvec(M_{green}-M)\times r_{green}-M]$
$\dbinom{0}{3/2}+\dfrac{\dbinom{0}{3/2}-\dbinom{x}{y}}{\sqrt{(0-x)^{2}+(3/2-y)^{2}}}\times \dfrac{3}{2}-M$ $\Rightarrow $ $d_{red}=\sqrt{\left(-x-\dfrac{3x}{\sqrt{4x^{2}+(3-2y)^{2}}}\right)^{2}+\left(3/2-y+\dfrac{9-6y}{2\sqrt{4x^{2}+(3-2y)^{2}}}\right)^{2}}$
$\dbinom{2}{0}+\dfrac{\dbinom{2}{0}-\dbinom{x}{y}}{\sqrt{(2-x)^{2}+(0-y)^{2}}}\times 2-M$ $\Rightarrow $ $d_{blue}=\sqrt{\left(2-x+\dfrac{4-2x}{\sqrt{(-2+x)^{2}+y^{2}}}\right)^{2}+\left(-y-\dfrac{2y}{\sqrt{(-2+x)^{2}+y^{2}}}\right)^{2}}$
$\dbinom{2}{3/2}+\dfrac{\dbinom{2}{3/2}-\dbinom{x}{y}}{\sqrt{(2-x)^{2}+(3/2-y)^{2}}}\times \dfrac{5}{2}-M$ $\Rightarrow $ $d_{red}=\sqrt{\left(2-x-\dfrac{5(-2+x)}{\sqrt{25+4x(-4+x)+4y(-3+y)}}\right)^{2}+\left(3/2-y+\dfrac{5(3/2-y)}{2\sqrt{(-2+x)^{2}+(-3/2+y)^{2}}}\right)^{2}}$
Setting these 3 equations to r and solving for x,y,r of the black circle results in:
$x=\dfrac{36}{23} , y=\dfrac{24}{23} , r=\dfrac{72}{23}$