The points A and B have coordinates (−2, 11) and (8, 1), respectively - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

The equation of a circle with center (h, k) and radius r is: $(x - h)^2 + (y - k)^2 = r^2$ Here, the center is (3, 6). We need to find the radius r.

The radius is the distance from the center to point A (or B). Using the distance formula: $r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Calculating the distance from (3, 6) to (-2, 11): $r = \sqrt{(3 - (-2))^2 + (6 - 11)^2} = \sqrt{(3 + 2)^2 + (6 - 11)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$

Thus, the equation of C is: $(x - 3)^2 + (y - 6)^2 = (5\sqrt{2})^2 = 50$

To verify if the point (10, 7) lies on the circle, we substitute x = 10 and y = 7 into the equation of the circle: $(10 - 3)^2 + (7 - 6)^2 = 50$ Calculating: $7^2 + 1^2 = 50 \Rightarrow 49 + 1 = 50 \Rightarrow 50 = 50$ Since the equation is satisfied, the point (10, 7) lies on the circle C.

The slope of the radius at the point (10, 7) can be found using the coordinates of the center (3, 6): $m_{radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 6}{10 - 3} = \frac{1}{7}$

The slope of the tangent line is the negative reciprocal: $m_{tangent} = -\frac{1}{m_{radius}} = -7$

Using the point-slope formula for the tangent line: $y - y_1 = m(x - x_1) \Rightarrow y - 7 = -7(x - 10)$ Expanding this gives: $y - 7 = -7x + 70 \Rightarrow y = -7x + 77$ Thus, the equation of the tangent line at the point (10, 7) is: $y = -7x + 77$.