Key Concepts and Formulas

• Angle in a Semicircle Theorem: An angle inscribed in a semicircle is a right angle. If PR is the diameter of a circle and Q is a point on the circle, then $\angle PQR = 90^{\circ}$.
• Midpoint Formula: The midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. This gives the center of the circle if P and R are endpoints of a diameter.
• Slope of a Line: The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$. The slopes of perpendicular lines satisfy $m_1m_2 = -1$.
• Equation of a Tangent: The tangents at the endpoints of radii to points Q and R intersect at S. This point S is external to the circle.

Step-by-Step Solution

Step 1: Use the angle in a semicircle theorem. Since PR is the diameter and Q lies on the circle, $\angle PQR = 90^\circ$. This means that PQ and QR are perpendicular.

Step 2: Calculate the slopes of PQ and QR. The slope of PQ is $m_{PQ} = \frac{10 - 2}{9 - (-3)} = \frac{8}{12} = \frac{2}{3}$. The slope of QR is $m_{QR} = \frac{4 - 10}{\alpha - 9} = \frac{-6}{\alpha - 9}$.

Step 3: Apply the perpendicularity condition. Since PQ and QR are perpendicular, $m_{PQ} \cdot m_{QR} = -1$. Therefore, $\frac{2}{3} \cdot \frac{-6}{\alpha - 9} = -1$ $\frac{-12}{3(\alpha - 9)} = -1$ $-12 = -3(\alpha - 9)$ $4 = \alpha - 9$ $\alpha = 13$ So, the coordinates of R are (13, 4).

Step 4: Find the center of the circle. The center of the circle is the midpoint of PR. The coordinates of P are (-3, 2) and the coordinates of R are (13, 4). The center is: $\left(\frac{-3 + 13}{2}, \frac{2 + 4}{2}\right) = \left(\frac{10}{2}, \frac{6}{2}\right) = (5, 3)$

Step 5: Determine the equations of the tangents at Q and R. The tangent at Q is perpendicular to the radius CQ, where C is the center (5, 3). The slope of CQ is $m_{CQ} = \frac{10 - 3}{9 - 5} = \frac{7}{4}$. The slope of the tangent at Q is $m_{tan,Q} = -\frac{4}{7}$. The equation of the tangent at Q is: $y - 10 = -\frac{4}{7}(x - 9)$ $7(y - 10) = -4(x - 9)$ $7y - 70 = -4x + 36$ $4x + 7y = 106$

The tangent at R is perpendicular to the radius CR. The slope of CR is $m_{CR} = \frac{4 - 3}{13 - 5} = \frac{1}{8}$. The slope of the tangent at R is $m_{tan,R} = -8$. The equation of the tangent at R is: $y - 4 = -8(x - 13)$ $y - 4 = -8x + 104$ $8x + y = 108$

Step 6: Find the intersection point S of the two tangents. We need to solve the system of equations: $4x + 7y = 106$ $8x + y = 108$ Multiply the second equation by -7: $-56x - 7y = -756$ Add this to the first equation: $(4x - 56x) + (7y - 7y) = (106 - 756)$ $-52x = -650$ $x = \frac{650}{52} = \frac{325}{26} = \frac{25}{2}$ Substitute $x = \frac{25}{2}$ into $8x + y = 108$: $8\left(\frac{25}{2}\right) + y = 108$ $100 + y = 108$ $y = 8$ So, the coordinates of S are $\left(\frac{25}{2}, 8\right)$.

Step 7: Use the given line equation to find k. S lies on the line $2x - ky = 1$. Substitute the coordinates of S into the equation: $2\left(\frac{25}{2}\right) - k(8) = 1$ $25 - 8k = 1$ $24 = 8k$ $k = 3$

Common Mistakes & Tips

• Double-check the slope calculations to avoid errors.
• Remember that the product of slopes of perpendicular lines is -1.
• Be careful when solving the system of linear equations.

Summary

We used the angle in a semicircle theorem to find the x-coordinate of point R. Then, we found the center of the circle as the midpoint of PR. We determined the equations of the tangents at points Q and R and solved for their intersection point S. Finally, we used the fact that S lies on the given line to solve for k.

The final answer is \boxed{3}.