The circle C has centre A(2,1) and passes through the point B(10, 7) - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 3

To find the equation of the tangent line l₁ at point B(10, 7), we need the gradient of the radius OB, where O is the center A(2, 1):

The gradient of line AB is:

$\text{slope}_{AB} = \frac{7 - 1}{10 - 2} = \frac{6}{8} = \frac{3}{4}$

The gradient of the tangent l₁ is the negative reciprocal of the radius gradient:

$\text{slope}_{l_1} = -\frac{4}{3}$

Using the point-slope form of a line,

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

Substituting the coordinates of point B(10, 7):

$y - 7 = -\frac{4}{3}(x - 10)$

Rearranging gives:

$y = -\frac{4}{3}x + \frac{40}{3} + 7\ \\ = -\frac{4}{3}x + \frac{40}{3} + \frac{21}{3} = -\frac{4}{3}x + \frac{61}{3}$

Thus, the equation of tangent line l₁ is:

$y = -\frac{4}{3}x + \frac{61}{3}$

To find the length of line segment PQ, we first need to find the coordinates of points P and Q where line l₂ intersects the circle C. Since l₂ is parallel to l₁, it will have the same gradient, which is -\frac{4}{3}.

Let M be the mid-point of AB:

$M = \left( \frac{2 + 10}{2}, \frac{1 + 7}{2} \right) = \left( 6, 4 \right)$

Using the point-slope form for line l₂,

$y - 4 = -\frac{4}{3}(x - 6)$

Now simplify:

$y = -\frac{4}{3}x + 8 + 4 = -\frac{4}{3}x + 12$

Now we set the equations for circle C and line l₂ equal to each other:

From circle C: $(x - 2)^2 + (y - 1)^2 = 100$

Substituting ( y = -\frac{4}{3}x + 12 ): $(x - 2)^2 + \left(-\frac{4}{3}x + 12 - 1\right)^2 = 100$

Solving for x will give us the x-coordinates of points P and Q. After finding both points, we utilize the distance formula:

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

Finally, we simplify the expression to obtain the length PQ in its simplest form.