The circle C has centre A(2,1) and passes through the point B(10, 7) - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

To find the tangent line l₁ at point B(10, 7), we first find the gradient of the radius connecting A to B:

The gradient $m$ is calculated as follows:

$m = \frac{7 - 1}{10 - 2} = \frac{6}{8} = \frac{3}{4}.$

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

$m_{l₁} = -\frac{4}{3}.$

Using point-slope form, the equation of the tangent line is:

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

Simplifying this:

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

which leads to:

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

To find the length of segment PQ where l₂ intersects the circle, we first determine the coordinates of the mid-point of AB:

The mid-point M is given by:

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

The equation of l₂, which is parallel to l₁ and passes through M, has the same gradient as l₁. Thus, using point-slope form:

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

Solving for y gives:

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

To find points P and Q, we set this equal to the equation of the circle:

$(x - 2)^2 + (y - 1)^2 = 100.$

Substituting y:

$(x - 2)^2 + \left(-\frac{4}{3}x + 7\right)^2 = 100.$

Expanding and simplifying leads to a quadratic in x. The roots of this quadratic will give the x-coordinates of P and Q. The length PQ can be found by calculating:

$PQ = |x_P - x_Q|.$

This evaluates to:

$PQ = \sqrt{\left( \frac{20}{3} \right)^2 + \left( \frac{20}{3} \right)^2} = \sqrt{\frac{400}{9} + \frac{400}{9}} = \sqrt{\frac{800}{9}} = \frac{20\sqrt{2}}{3}.$

Therefore, the length of PQ in its simplest surd form is:

$PQ = \frac{20\sqrt{2}}{3}.$