Radius of Incircle of Triangle PQR

To find the radius of the incircle of the triangle $PQR$, we consider the given configuration from the passage:

The given curves are the circle:
$x^2 + y^2 = 9$
and the parabola:
$y^2 = 8x$

Step 1: Finding the Points of Intersection $P$ and $R$

Substituting $y^2 = 8x$ into the equation of the circle:
$x^2 + 8x - 9 = 0$
$(x + 9)(x - 1) = 0$

Since $x \ge 0$ for points on the parabola $y^2 = 8x$, we take:
$x = 1$

Then:
$y^2 = 8(1) = 8 \implies y = \pm 2\sqrt{2}$

Thus, the points of intersection in the first and fourth quadrants are:
$P = (1, 2\sqrt{2}) \quad \text{and} \quad R = (1, -2\sqrt{2})$

Step 2: Finding the Coordinates of Point $Q$

The equation of the tangent to the circle $x^2 + y^2 = 9$ at the point $P(1, 2\sqrt{2})$ is:
$x(1) + y(2\sqrt{2}) = 9 \implies x + 2\sqrt{2}y = 9$

The tangent intersects the $x$-axis at $Q$ (where $y = 0$):
$x + 2\sqrt{2}(0) = 9 \implies x = 9$

Thus, the coordinates of $Q$ are:
$Q = (9, 0)$

(By symmetry, the tangent to the circle at $R(1, -2\sqrt{2})$ also intersects the $x$-axis at $Q(9, 0)$).

Step 3: Determining the Side Lengths of $\triangle PQR$

Using the distance formula, the lengths of the sides of $\triangle PQR$ are:

• $PR = |2\sqrt{2} - (-2\sqrt{2})| = 4\sqrt{2}$
• $PQ = \sqrt{(9 - 1)^2 + (0 - 2\sqrt{2})^2} = \sqrt{64 + 8} = \sqrt{72} = 6\sqrt{2}$
• $QR = \sqrt{(9 - 1)^2 + (0 - (-2\sqrt{2}))^2} = \sqrt{64 + 8} = \sqrt{72} = 6\sqrt{2}$

Step 4: Calculating the Semi-Perimeter and Area of $\triangle PQR$

The semi-perimeter $s$ is:
$s = \frac{PQ + QR + PR}{2} = \frac{6\sqrt{2} + 6\sqrt{2} + 4\sqrt{2}}{2} = \frac{16\sqrt{2}}{2} = 8\sqrt{2}$

The triangle is symmetric with respect to the $x$-axis, with base $PR$ lying along the vertical line $x = 1$. The altitude from the vertex $Q(9, 0)$ to the base $PR$ is:
$h = 9 - 1 = 8$

Thus, the area of $\triangle PQR$ is:
$\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\sqrt{2} \times 8 = 16\sqrt{2}$

Step 5: Finding the Inradius $r$

The radius of the incircle is given by the formula:
$r = \frac{\Delta}{s} = \frac{16\sqrt{2}}{8\sqrt{2}} = 2$

Therefore, the radius of the incircle of $\triangle PQR$ is $2$.

Correct Option: (D)