A circle has equation (x - 1)^2 + (y - 2)^2 = 25. A line has equation 3x + y = 20. (a) Find the coordinates of the two points at which the line intersects the circle. [6] (b) Find the exact length of the chord joining these two points of intersection. [2]

AS Level Maths 9709: Intersection of a Line and a Circle

Rearrange the line for substitution. From $3x + y = 20$ :

$y = 20 - 3x.$

Substitute into the circle’s equation. In $(x-1)^2 + (y-2)^2 = 25$ , replace $y$ :

$(x - 1)^2 + (20 - 3x - 2)^2 = 25 \implies (x-1)^2 + (18 - 3x)^2 = 25.$

Expand each bracket.

$(x-1)^2 = x^2 - 2x + 1$

$(18 - 3x)^2 = 324 - 108x + 9x^2$

Adding these together:

$x^2 - 2x + 1 + 9x^2 - 108x + 324 = 25$

$10x^2 - 110x + 325 = 25$

$10x^2 - 110x + 300 = 0.$

Solve the quadratic. Dividing through by $10$ :

$x^2 - 11x + 30 = 0 \implies (x - 5)(x - 6) = 0 \implies x = 5 \text{ or } x = 6.$

Find the corresponding $y$ -coordinates, using $y = 20 - 3x$ :

Check both points lie on the circle:

$(5-1)^2 + (5-2)^2 = 16 + 9 = 25. \checkmark \qquad (6-1)^2 + (2-2)^2 = 25 + 0 = 25. \checkmark$

So the line meets the circle at $\boxed{(5, 5)}$ and $\boxed{(6, 2)}$ .

Using the distance formula between $(5, 5)$ and $(6, 2)$ :

$\text{length} = \sqrt{(6 - 5)^2 + (2 - 5)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}.$

Since $10 = 2 \times 5$ has no square factor greater than $1$ , $\sqrt{10}$ is already in its simplest surd form.

So the chord has exact length $\boxed{\sqrt{10}}$ .