JEE Main 2025 — Mathematics Coordinate Geometry

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

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

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

$$(x-a)^2+(y-r)^2=r^2$$(x−a)2+(y−r)2=r2

satsty $\mathrm{P}(4,6) \Rightarrow \mathrm{a}^2-8 \mathrm{a}-12 r+52=0 \ldots \ldots$P(4,6)⇒a2−8a−12r+52=0…… (2)

From equation (1)

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

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