Step 1: Find the intersection points of the circle and the parabola.

The equation of the circle is $x^2 + y^2 = 9$. The equation of the parabola is $y^2 = 8x$.

To find the intersection points, substitute the expression for $y^2$ from the parabola's equation into the circle's equation: $x^2 + 8x = 9$ $x^2 + 8x - 9 = 0$ Factoring the quadratic equation, we get: $(x+9)(x-1) = 0$ This gives two possible values for x: $x = -9$ and $x = 1$.

From the parabola equation, $y^2 = 8x$, we know that $x$ must be non-negative ($x \ge 0$) for $y$ to be a real number. Therefore, we must choose $x=1$.

Now, substitute $x=1$ back into the parabola's equation to find the y-coordinates: $y^2 = 8(1) = 8$ $y = \pm \sqrt{8} = \pm 2\sqrt{2}$

The problem states that P is in the first quadrant and Q is in the fourth quadrant. Thus, the coordinates are: $P = (1, 2\sqrt{2})$ $Q = (1, -2\sqrt{2})$

Step 2: Find the coordinates of point R.

R is the intersection point of the tangents to the circle at P and Q with the x-axis. Due to symmetry of points P and Q with respect to the x-axis, the tangents will intersect the x-axis at the same point.

The equation of the tangent to the circle $x^2 + y^2 = r^2$ at a point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. For our circle, $r^2 = 9$. The tangent at $P(1, 2\sqrt{2})$ is: $x(1) + y(2\sqrt{2}) = 9$ $x + 2\sqrt{2}y = 9$ To find the intersection with the x-axis, we set $y=0$: $x + 2\sqrt{2}(0) = 9 \implies x=9$ So, the point R is $(9, 0)$.

Step 3: Find the coordinates of point S.

S is the intersection point of the tangents to the parabola at P and Q with the x-axis. Again, due to symmetry, both tangents will intersect the x-axis at the same point.

The equation of the tangent to the parabola $y^2 = 4ax$ at a point $(x_1, y_1)$ is $yy_1 = 2a(x+x_1)$. For our parabola, $y^2=8x$, so $4a=8$, which means $a=2$. The tangent at $P(1, 2\sqrt{2})$ is: $y(2\sqrt{2}) = 2(2)(x+1)$ $2\sqrt{2}y = 4(x+1)$ To find the intersection with the x-axis, we set $y=0$: $2\sqrt{2}(0) = 4(x+1) \implies 0 = 4(x+1) \implies x = -1$ So, the point S is $(-1, 0)$.

Step 4: Calculate the circumradius of triangle PRS.

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

We can calculate the lengths of the sides of the triangle:

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

The area of the triangle PRS, denoted by $\Delta$, can be calculated using the base RS on the x-axis and the height as the y-coordinate of P. Base $RS = 10$. Height = $y_P = 2\sqrt{2}$. $\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 2\sqrt{2} = 10\sqrt{2}$

The radius of the circumcircle, $R_c$, is given by the formula $R_c = \frac{abc}{4\Delta}$, where $a, b, c$ are the side lengths. $R_c = \frac{(PR)(PS)(RS)}{4\Delta} = \frac{(6\sqrt{2})(2\sqrt{3})(10)}{4(10\sqrt{2})}$ $R_c = \frac{120\sqrt{6}}{40\sqrt{2}}$ We can simplify by cancelling common terms: $R_c = \frac{6\sqrt{2} \cdot 2\sqrt{3} \cdot 10}{4 \cdot 10\sqrt{2}} = \frac{6 \cdot 2\sqrt{3}}{4} = \frac{12\sqrt{3}}{4} = 3\sqrt{3}$

The radius of the circumcircle of triangle PRS is $3\sqrt{3}$ units.

Step 5: Compare the result with the given options.

The calculated radius is $3\sqrt{3}$ units. This corresponds to option B.

A: 5 units B: 3 $\sqrt3$ units C: 3 $\sqrt2$ units D: 2 $\sqrt3$ units

The correct option is B.