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


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:
$x=\dfrac{36}{23} , y=\dfrac{24}{23} , r=\dfrac{72}{23}$
