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

If the normals of the parabola y² = 4x drawn at the end points of its latus rectum are tangents to the circle (x - 3)² + (y + 2)² = r², then the value of r² is $\text{Given: parabola } y^2 = 4x$ $\text{circle } (x-3)^2 + (y+2)^2 = r^2$ $\text{normals at ends of latus rectum are tangents to the circle}$ $\text{To find: } r^2$ $y^2 = 4x \quad\Rightarrow\quad 4a = 4 \Rightarrow a = 1$ $\text{opens right, vertex at the origin}$ $\text{Focus } S = (1, 0)$ $\text{Latus rectum: } x = 1, \text{ length } 4a = 4$ $\text{End points } P(1, 2), \ Q(1, -2)$ $\text{Circle: centre } C(3, -2), \text{ radius } r$ $\text{Normal at } (at^2, 2at): \ y = -tx + 2at + at^3$ $a = 1 \Rightarrow y = -tx + 2t + t^3$ $P(1, 2): 2t = 2 \Rightarrow t = 1$ $Q(1, -2): 2t = -2 \Rightarrow t = -1$ $\text{At } P:\ y = -x + 2 + 1 = -x + 3$ $\Rightarrow x + y = 3$ $\text{At } Q:\ y = x - 2 - 1 = x - 3$ $\Rightarrow x - y = 3$ $\text{Tangency: distance from } C(3,-2) \text{ to each line} = r$ $d = \dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$ $x + y - 3 = 0:\quad d = \dfrac{|3 - 2 - 3|}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$ $x - y - 3 = 0:\quad d = \dfrac{|3 + 2 - 3|}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$ $r = \sqrt{2}$ $\boxed{r^2 = 2}$