Circle Equation from Centre and a Point

Given a circle’s centre and one point on it, the radius is just the distance between them. Compute $r^2$ directly with the distance formula (no need to take a root), then drop it into the standard form.

A circle has centre $C(2, -1)$ and passes through the point $P(5, 3)$. Find the equation of the circle. [4]

The working

Step 1, find $r^2$ using the distance from $C$ to $P$ (keep it squared): $r^2 = (5 - 2)^2 + (3 - (-1))^2 = 3^2 + 4^2 = 9 + 16 = 25 \quad \text{(M1, A1)}$

Step 2, substitute the centre $(2, -1)$ and $r^2 = 25$ into $(x - a)^2 + (y - b)^2 = r^2$: $(x - 2)^2 + (y + 1)^2 = 25 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Compute $r^2$ directly ($= 25$); don’t find $r = 5$ and then square it back, more steps, more chances to slip.
• Mind the signs in the brackets: centre $(2, -1)$ gives $(x - 2)$ and $(y - (-1)) = (y + 1)$.
• Watch the double negative in the distance: $3 - (-1) = 4$.

Common mistakes

• Writing $r = 25$ instead of $r^2 = 25$ on the right-hand side.
• Sign slip: $(y - 1)$ instead of $(y + 1)$ for centre $y = -1$.
• $(3 - (-1))^2 = 2^2$ (dropping the double negative).

Full method: Equation of a Circle notes. Topic home: Circle Geometry pillar.