The points \( A \) and \( B \) have coordinates \( (-2,11) \) and \( (8,1) \) respectively. Given that \( A B \) is a diameter of the circle \( C \) (a) show that the centre of \( C \) has coordinates \( (3,6) \), (b) find an equation for \( C \). (c) Verify that the point \( (10,7) \) lies on \( C \). (d) Find an equation of the tangent to \( C \) at the point \( (10,7) \), giving your answer in the form \( y= \) \( m x+c \), where \( m \) and \( c \) are constants.

To solve the problem, we will go through each part step by step. ### Part (a): Show that the centre of \( C \) has coordinates \( (3,6) \). The center of the circle \( C \) is the midpoint of the diameter \( AB \). The coordinates of points \( A \) and \( B \) are given as \( A(-2, 11) \) and \( B(8, 1) \). The midpoint \( M \) of segment \( AB \) can be calculated using the midpoint formula: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] Substituting the coordinates of \( A \) and \( B \): \[ M = \left( \frac{-2 + 8}{2}, \frac{11 + 1}{2} \right) = \left( \frac{6}{2}, \frac{12}{2} \right) = (3, 6) \] Thus, the center of circle \( C \) is \( (3, 6) \). ### Part (b): Find an equation for \( C \). To find the equation of the circle, we need the radius. The radius is half the length of the diameter \( AB \). We first calculate the distance \( AB \) using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substituting the coordinates of \( A \) and \( B \): \[ AB = \sqrt{(8 - (-2))^2 + (1 - 11)^2} = \sqrt{(8 + 2)^2 + (1 - 11)^2} = \sqrt{10^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \] The radius \( r \) is half of this distance: \[ r = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \] The standard equation of a circle with center \( (h, k) \) and radius \( r \) is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Substituting \( (h, k) = (3, 6) \) and \( r = 5\sqrt{2} \): \[ (x - 3)^2 + (y - 6)^2 = (5\sqrt{2})^2 = 50 \] Thus, the equation of circle \( C \) is: \[ (x - 3)^2 + (y - 6)^2 = 50 \] ### Part (c): Verify that the point \( (10,7) \) lies on \( C \). 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 the left side: \[ (7)^2 + (1)^2 = 49 + 1 = 50 \] Since the left side equals the right side, the point \( (10, 7) \) lies on circle \( C \). ### Part (d): Find an equation of the tangent to \( C \) at the point \( (10,7) \). To find the equation of the tangent line at the point \( (10, 7) \), we first need the slope of the radius at that point. The slope of the radius from the center \( (3, 6) \) to the point \( (10, 7) \) is: \[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 6}{10 - 3} = \frac{1}{7} \] The slope of the tangent line is the negative reciprocal of the slope of the radius: \[ \text{slope of tangent} = -\frac{1}{\frac{1}{7}} = -7 \] Now, we can use the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \] Substituting \( m = -7 \) and the point \( (10, 7) \): \[ y -