A circle C has centre M (6, 4) and radius 3 - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2

To find the angle TMQ, we need to determine the coordinates of the points T and Q along with point P (12, 6).

1. Calculate the slope of line MP:

   • The coordinates of M are (6, 4) and P are (12, 6).
   • The slope, $m_{MP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 4}{12 - 6} = \frac{2}{6} = \frac{1}{3}$.

2. The tangent at point T, which lies on the circle, is perpendicular to MP, so:

   • $m_{TQ} = -\frac{1}{m_{MP}} = -3$.

3. Using the coordinates of T and Q (which can be calculated), the angle $heta$ can be found using:

   • $\theta = \tan^{-1}\left|\frac{m_{TMP} - m_{TQ}}{1 + m_{TMP} \cdot m_{TQ}}\right|$

4. By substituting the respective slopes, we can derive the tangent and then calculate the angle TMQ, confirming it equals approximately 1.0766 radians.

The shaded region TPQ can be computed by finding the area of triangle TPQ and subtracting the area of the sector formed by the circle.

1. Area of Triangle TPQ:

   • Use the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
   • The base TP can be calculated from the coordinates of points T and P.
   • And the height is the perpendicular distance from Q to line TP.

2. Area of the Sector:

   • The angle at the center M corresponding to the arc TQ can be calculated based on the angle TMQ.
   • Then, using the radius, the area of the sector is: $\text{Area}_{sector} = \frac{1}{2} r^2 \theta$ where $heta$ is in radians.

3. Final Area Calculation:

   • The area of the shaded region is: $\text{Area}_{shaded} = \text{Area}_{triangle} - \text{Area}_{sector}$
   • Substitute the values to arrive at the final answer accurate to three decimal places.