Circumcircle Radius of Triangle Formed by Tangent Intersections

To find the radius of the circumcircle of triangle $PRS$, we first determine the coordinates of the points $P$, $R$, and $S$.

Step 1: Find the Points of Intersection $P$ and $Q$

The given equations are:

1. Circle: $x^2 + y^2 = 9$
2. Parabola: $y^2 = 8x$

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

Since $y^2 \ge 0$, we have $x \ge 0$, which gives:
$x = 1$

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

Since $P$ is in the first quadrant and $Q$ is in the fourth quadrant:
$P = (1, 2\sqrt{2}) \quad \text{and} \quad Q = (1, -2\sqrt{2})$

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Step 2: Find the Coordinates of Point $R$

The equation of the tangent to the circle $x^2 + y^2 = 9$ at the point $P(x_1, y_1) = (1, 2\sqrt{2})$ is:
$x x_1 + y y_1 = 9 \implies x + 2\sqrt{2}y = 9$

The tangent intersects the $x$-axis at $R$. Setting $y = 0$:
$x = 9 \implies R = (9, 0)$

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Step 3: Find the Coordinates of Point $S$

The equation of the tangent to the parabola $y^2 = 8x$ at the point $P(x_1, y_1) = (1, 2\sqrt{2})$ is:
$y y_1 = 4(x + x_1) \implies 2\sqrt{2}y = 4(x + 1)$

The tangent intersects the $x$-axis at $S$. Setting $y = 0$:
$0 = 4(x + 1) \implies x = -1 \implies S = (-1, 0)$

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Step 4: Find the Circumradius of Triangle $PRS$

The vertices of triangle $PRS$ are:
$P(1, 2\sqrt{2}), \quad R(9, 0), \quad S(-1, 0)$

The lengths of the sides of $\triangle PRS$ are:

• $RS = 9 - (-1) = 10$
• $PR = \sqrt{(9 - 1)^2 + (0 - 2\sqrt{2})^2} = \sqrt{64 + 8} = \sqrt{72} = 6\sqrt{2}$
• $PS = \sqrt{(-1 - 1)^2 + (0 - 2\sqrt{2})^2} = \sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3}$

The area ($\Delta$) of $\triangle PRS$, using base $RS$ on the $x$-axis and height equal to the $y$-coordinate of $P$, is:
$\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 2\sqrt{2} = 10\sqrt{2}$

Using the circumradius formula $R_{\text{circum}} = \frac{abc}{4\Delta}$:
$R_{\text{circum}} = \frac{(6\sqrt{2}) \times (2\sqrt{3}) \times 10}{4 \times 10\sqrt{2}} = \frac{120\sqrt{6}}{40\sqrt{2}} = 3\sqrt{3}$

Thus, the radius of the circumcircle of triangle $PRS$ is $3\sqrt{3}$.

Correct Answer: (B)