Let $r$ be the radius of the circle, which touches x -axis at point $(\mathrm{a}, 0), \mathrm{a} \lt 0$ and the parabola $\mathrm{y}^2=9 \mathrm{x}$ at the point $(4,6)$. Then $r$ is equal to ________

Solution

$\begin{aligned}<br/>& (x-a)^2+(y-r)^2=r^2 \\ & (4-a)^2+(6-r)^2=r^2 \\ & 16+a^2-8 a+36+r^2-12 r=r^2 \\ & a^2-8 a-12 r+52=0<br/>\end{aligned}$
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
$\begin{aligned}<br/>& \therefore \quad C P=r \\ & \frac{3 a-4 r+12}{5}= \pm r<br/>\end{aligned}$
$3 \mathrm{a}+12=4 \mathrm{r} \pm 5 \mathrm{r}\left\{\begin{array}{l}<br/>\mathrm{ar} \\ -\mathrm{r}<br/>\end{array}\right......(1)$
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 $r=30$

Hence, the correct answer is 30.