The points A and B have coordinates (–2, 11) and (8, 1), respectively.\n\nGiven that AB is a diameter of the circle C,\n\n(a) show that the centre of C has coordinates (3, 6).\n\n(b) find an equation for C.\n\n(c) Verify that the point (10, 7) lies on C.\n\n(d) Find an equation of the tangent to C at the point (10, 7), giving your answer in the form y = mx + c, where m and c are constants. - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 3

The general equation of a circle with center (h, k) and radius r is:

$(x - h)^2 + (y - k)^2 = r^2$

Here, the center (h, k) is (3, 6). First, we need to calculate the radius r, which is half of the diameter AB. We can find the length of AB using the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Calculating the distance AB:

$d = \sqrt{(8 - (-2))^2 + (1 - 11)^2} = \sqrt{(8 + 2)^2 + (1 - 11)^2} = \sqrt{10^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}$

Thus, the radius r = \frac{d}{2} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}.

Therefore, the equation of the circle C is:

$(x - 3)^2 + (y - 6)^2 = (5\sqrt{2})^2 = 50$

So, the equation for C is:

$(x - 3)^2 + (y - 6)^2 = 50$

To verify if the point (10, 7) lies on the circle C, we substitute x = 10 and y = 7 into the equation:

$(10 - 3)^2 + (7 - 6)^2 = 50$

Calculating:

$(7)^2 + (1)^2 = 50$

which simplifies to:

$49 + 1 = 50$

Since $50 = 50$, the point (10, 7) indeed lies on the circle C.

To find the equation of the tangent line at the point (10, 7), we need the gradient of the radius at this point. The radius runs from the center (3, 6) to the point (10, 7). Thus, the gradient m of the radius is:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 6}{10 - 3} = \frac{1}{7}$

The gradient of the tangent line is the negative reciprocal of the radius gradient:

$m_{tangent} = -\frac{1}{m} = -7$

Using the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Substituting in the point (10, 7) and the gradient -7:

$y - 7 = -7(x - 10)$

Simplifying this gives:

$y - 7 = -7x + 70$

thus,

$y = -7x + 77$

Therefore, the equation of the tangent to C at the point (10, 7) is:

y = -7x + 77.