$$\begin{aligned} & (x-a)^2+(y-r)^2=r^2 \\ & (4-a)^2+(6-r)^2=r^2 \\ & 16+a^2-8 a+36+r^2-12 r=r^2 \\ & a^2-8 a-12 r+52=0 \end{aligned}$$

Tangent to parabola at $(4,6)$ is

$$6.4=9 .\left(\frac{x+4}{2}\right) \text { i.e. } 3 x-4 y+12=0$$

This is also tangent to the circle

$$\begin{aligned} & \therefore \quad C P=r \\ & \frac{3 a-4 r+12}{5}= \pm r \end{aligned}$$

$3 \mathrm{a}+12=4 \mathrm{r} \pm 5 \mathrm{r}\left\{\begin{array}{l}\mathrm{ar} \\ -\mathrm{r}\end{array}\right.\quad\text{..... (1)}$

equation of circle is

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

satsty $P(4,6) \Rightarrow a^2-8 a-12 r+52=0\quad\text{..... (2)}$

From equation (1)

If $\mathrm{a}+4=3 \mathrm{r}$ then $\mathrm{a}=+6$ (rejected)

If $3 a+12=-r$ then $a=-14$ and $r=30$