Question:

Step 1: Determine the coordinates of point R using the diameter property. Since PR is the diameter of the circle, the angle subtended by PR at any point on the circumference, such as Q, is $90^\circ$. Therefore, the lines PQ and QR are perpendicular. The product of their slopes must be $-1$. The coordinates are $P(-3,2)$, $Q(9,10)$, and $R(\alpha,4)$. The slope of PQ, $m_{PQ}$, is: $$ m_{PQ} = \frac{10 - 2}{9 - (-3)} = \frac{8}{12} = \frac{2}{3} $$ The slope of QR, $m_{QR}$, is: $$ m_{QR} = \frac{4 - 10}{\alpha - 9} = \frac{-6}{\alpha - 9} $$ Since $PQ \perp QR$, we have $m_{PQ} \cdot m_{QR} = -1$: $$ \left(\frac{2}{3}\right) \cdot \left(\frac{-6}{\alpha - 9}\right) = -1 $$ $$ \frac{-12}{3(\alpha - 9)} = -1 $$ $$ \frac{-4}{\alpha - 9} = -1 $$ $$ 4 = \alpha - 9 $$ $$ \alpha = 13 $$ Thus, the coordinates of point R are $(13,4)$. Step 2: Find the equation of the circle C. Since PR is the diameter, the center of the circle is the midpoint of PR. Let the center be $M(h,k)$: $$ M = \left(\frac{-3 + 13}{2}, \frac{2 + 4}{2}\right) = \left(\frac{10}{2}, \frac{6}{2}\right) = (5,3) $$ The radius squared, $r^2$, can be found using the distance from the center M to P (or R). $$ r^2 = (5 - (-3))^2 + (3 - 2)^2 = (8)^2 + (1)^2 = 64 + 1 = 65 $$ The equation of the circle C is $(x-h)^2 + (y-k)^2 = r^2$: $$ (x-5)^2 + (y-3)^2 = 65 $$ Expanding this, we get: $$ x^2 - 10x + 25 + y^2 - 6y + 9 = 65 $$ $$ x^2 + y^2 - 10x - 6y - 31 = 0 $$ Step 3: Determine the coordinates of point S, the intersection of tangents. The point S is the intersection of the tangents to the circle C at points Q and R. The equation of the tangent to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ at a point $(x_1, y_1)$ is $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. For the circle $x^2 + y^2 - 10x - 6y - 31 = 0$, we have $g=-5$ and $f=-3$. Equation of the tangent at R$(13,4)$: $$ x(13) + y(4) - 5(x+13) - 3(y+4) - 31 = 0 $$ $$ 13x + 4y - 5x - 65 - 3y - 12 - 31 = 0 $$ $$ 8x + y - 108 = 0 \quad (\text{Equation 1}) $$ Equation of the tangent at Q$(9,10)$: $$ x(9) + y(10) - 5(x+9) - 3(y+10) - 31 = 0 $$ $$ 9x + 10y - 5x - 45 - 3y - 30 - 31 = 0 $$ $$ 4x + 7y - 106 = 0 \quad (\text{Equation 2}) $$ To find the coordinates of S, we solve Equation 1 and Equation 2. From Equation 1, $y = 108 - 8x$. Substitute this into Equation 2: $$ 4x + 7(108 - 8x) - 106 = 0 $$ $$ 4x + 756 - 56x - 106 = 0 $$ $$ -52x + 650 = 0 $$ $$ 52x = 650 $$ $$ x = \frac{650}{52} = \frac{325}{26} = \frac{25}{2} $$ Now, substitute the value of $x$ back into the expression for $y$: $$ y = 108 - 8\left(\frac{25}{2}\right) = 108 - 4(25) = 108 - 100 = 8 $$ So, the coordinates of point S are $\left(\frac{25}{2}, 8\right)$. Step 4: Use the condition that S lies on the given line to find k. The point $S\left(\frac{25}{2}, 8\right)$ lies on the line $2x-ky=1$. Substitute the coordinates of S into the equation of the line: $$ 2\left(\frac{25}{2}\right) - k(8) = 1 $$ $$ 25 - 8k = 1 $$ $$ 8k = 25 - 1 $$ $$ 8k = 24 $$ $$ k = \frac{24}{8} $$ $$ k = 3 $$ The final answer is $\boxed{3}$.