Radius of Circle Tangent to Normals of Parabola Latus Rectum

To find the value of $r^2$, we first determine the equations of the normal lines to the parabola at the endpoints of its latus rectum.

The given equation of the parabola is: $y^2 = 4x$

Comparing this with the standard form $y^2 = 4ax$, we get $a = 1$.

The focus of the parabola is at $(1, 0)$, and the endpoints of the latus rectum are: $P = (1, 2) \quad \text{and} \quad Q = (1, -2)$

Step 1: Finding the equations of the normals

Differentiating $y^2 = 4x$ with respect to $x$: $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$

• At point $P(1, 2)$: The slope of the tangent is $m_t = \frac{2}{2} = 1$. Therefore, the slope of the normal is $m_n = -1$.

  The equation of the normal at $P(1, 2)$ is: $y - 2 = -1(x - 1) \implies x + y - 3 = 0$

• At point $Q(1, -2)$: The slope of the tangent is $m_t = \frac{2}{-2} = -1$. Therefore, the slope of the normal is $m_n = 1$.

  The equation of the normal at $Q(1, -2)$ is: $y - (-2) = 1(x - 1) \implies x - y - 3 = 0$

Step 2: Using the tangency condition for the circle

The given circle is: $(x - 3)^2 + (y + 2)^2 = r^2$

The center of the circle is $C(3, -2)$ and its radius is $r$.

Since the normal lines are tangential to the circle, the perpendicular distance from the center $C(3, -2)$ to both lines must equal the radius $r$.

Calculating the perpendicular distance $d$ from $C(3, -2)$ to the line $x + y - 3 = 0$: $r = \frac{|1(3) + 1(-2) - 3|}{\sqrt{1^2 + 1^2}} = \frac{|3 - 2 - 3|}{\sqrt{2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$

Similarly, checking for the line $x - y - 3 = 0$: $r = \frac{|1(3) - 1(-2) - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|3 + 2 - 3|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Step 3: Calculate $r^2$

Squaring the radius $r = \sqrt{2}$: $r^2 = (\sqrt{2})^2 = 2$