Circles

From the picture we have one equation to solve and it is

$\sqrt{(r+3)^{2}-r^{2}}=2+\sqrt{(5-r)^{2}-r^{2}}$

$\sqrt{9+6r}=2+\sqrt{25-10r}$

$4r-3=\sqrt{9+6r}$

$16r^{2}-24r+9=9+6r,r=\frac{15}{8}$

Hence

$p=\boxed{8}$

And radius of circle can be known by rearrange form of equation that was given.

Relevant Wiki: Descartes' Circle Theorem

By some calculation, we will know that two known circles' radius are 3 and 5,

because the wanted circle is tangent to the red line, so it is sufficient to clone the circle, as the figure above. It will do a great help. (Also sorry for the figure, it should has only one cloned circle.)

let $k_1=\frac{1}{3}, k_2=-\frac{1}{5}, k_3=k_4=\frac{1}{r}$ , the sign of k's are respect to if it where externally tangent or internally tangent to each other.

$2(k_1^2+k_2^2+k_3^2+k_4^2)=(k_1+k_2+k_3+k_4)^2$

$2(\frac{1}{9}+\frac{1}{25}+\frac{2}{r^2})=(\frac{1}{3}-\frac{1}{5}+\frac{2}{r})^2$

Solving this, yield $r = \boxed{\frac{15}{8}}$ .