JEE Main 2025 — Parabola Question with Solution

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

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

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

This is also tangent to the circle

$$\begin{array}{rl}& \therefore CP=r\\ & \frac{3a-4r+12}{5}=\pm r\end{array}$$

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

equation of circle is

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

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

From equation (1)

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

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