Coordinate Geometry: Question 3

Worked solution

Part (a): Centre and radius

Method 1. General form. The equation $x^2 + y^2 + 2gx + 2fy + c = 0$ has centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$. Comparing with $x^2 + y^2 - 4x + 8y + 11 = 0$:

$2g = -4 \implies g = -2, \qquad 2f = 8 \implies f = 4, \qquad c = 11.$

So the centre is $(-g, -f) = (2, -4)$, and the radius is

$r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + 4^2 - 11} = \sqrt{4 + 16 - 11} = \sqrt{9} = 3.$

Method 2, completing the square (check). Group the $x$ and $y$ terms:

$(x^2 - 4x) + (y^2 + 8y) + 11 = 0$

$(x - 2)^2 - 4 + (y + 4)^2 - 16 + 11 = 0$

$(x - 2)^2 + (y + 4)^2 = 9.$

This confirms centre $(2, -4)$ and radius $\sqrt{9} = 3$, the same as Method 1.

So the centre is $\boxed{(2, -4)}$ and the radius is $\boxed{3}$.

Part (b): Showing $M(5, -4)$ lies on the circle

The distance from the centre $C(2, -4)$ to $M(5, -4)$ is

$CM = \sqrt{(5 - 2)^2 + (-4 - (-4))^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3.$

Since $CM = 3$, which is exactly equal to the radius found in part (a), the point $M$ is at distance $3$ from the centre, so $M$ lies on the circle. $\blacksquare$

Part (c): Tangent to the circle at $M$

A tangent to a circle is always perpendicular to the radius drawn to the point of contact, so we first need the gradient of $CM$.

$\text{gradient of } CM = \frac{-4 - (-4)}{5 - 2} = \frac{0}{3} = 0.$

The radius $CM$ is horizontal (its gradient is $0$), since $C(2,-4)$ and $M(5,-4)$ share the same $y$-coordinate.

A line perpendicular to a horizontal line is vertical. Since the tangent passes through $M(5, -4)$, its equation is

$x = 5.$

Final answers

• (a) Centre $(2, -4)$, radius $3$
• (b) $CM = 3 =$ radius, so $M$ lies on the circle
• (c) Tangent at $M$: $x = 5$