If the normals of the parabola $y^2=4x$ drawn at the end points of its latus rectum are tangents to the circle ${\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}+2\right)}^2={\mathrm{r}}^2$, then the value of r² is

$$\begin{gathered} \mathrm{Given} {\mathrm{y}}^2=4{\mathrm{x}}_{ } \\ \Rightarrow a=1 \\ Now \mathrm{L}=(\mathrm{a},2\mathrm{a})=(1,2) \\ \mathrm{Eq}uation \mathrm{of} \mathrm{the} \mathrm{normal} at L is \\ \mathrm{y}-{\mathrm{y}}_1=\frac{-{\mathrm{y}}_1}{2\mathrm{a}}(\mathrm{x}-{\mathrm{x}}_1) \\ \Rightarrow \mathrm{y}-2=\frac{-2}{2}(\mathrm{x}-1) \\ \Rightarrow \mathrm{y}-2=-\mathrm{x}+1 \\ \Rightarrow \mathrm{x}+\mathrm{y}-3=0 \end{gathered}$$

it is tangent of the circle

 $$\begin{gathered} {(\mathrm{x}-3)}^2+{(\mathrm{y}+2)}^2={\mathrm{r}}^2 with centre\left(C\right)=(3,-2) then \mathrm{r}=\mathrm{d} \\ \Rightarrow \mathrm{r}=\frac{\left|\left(\cancel{3}-2-\cancel{3}\right)\right|}{\sqrt{2}}=\sqrt{2} \\ \Rightarrow {\mathrm{r}}^2=2 \end{gathered}$$