JEE Main 2024 — 8 April, Shift 2 — Question 1

Image of point $(-4,5)$

$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=-2\left(\frac{a x_{1}+b y_{1}+c}{a^{2}+b^{2}}\right)$

Line: $x+2 y-2=0$

$\begin{aligned} \frac{x+4}{1}=\frac{y-5}{2} & =-2\left(\frac{-4+10-2}{1^{2}+2^{2}}\right) & =\frac{-8}{5} \end{aligned}$

$x=-4-\frac{8}{5}=-\frac{28}{5}$

$y=-\frac{16}{5}+5=\frac{9}{5}$

Point lies on circle $(x+4)^{2}+(y-3)^{2}=r^{2}$

$\frac{64}{25}+\left(\frac{9}{5}-3\right)^{2}=\mathrm{r}^{2}$

$\frac{100}{25}=\mathrm{r}^{2}, \mathrm{r}=2$