Let r be the radius of the circle, which touches x -axis at point...

$$\begin{aligned}& (x-a)^2+(y-r)^2=r^2 \\ & (4-a)^2+(6-r)^2=r^2 \\ & 16+a^2-8 a+36+r^2-12 r=r^2 \\ & a^2-8 a-12 r+52=0\end{aligned}$$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......(1)$equation of circle is$(x-a)^2+(y-r)^2=r^2$satsty$\mathrm{P}(4,6) \Rightarrow \mathrm{a}^2-8 \mathrm{a}-12 r+52=0 \ldots \ldots$(2)From equation (1)$If \mathrm{a}+4=3 \mathrm{r} then \mathrm{a}=$+6$ (rejected)If 3 a+12=-r$then$a=-14 and r=30$