Midpoints & Perpendicular Bisectors
Every point on a circle is the same distance from its centre — and that single fact means the centre always sits on the perpendicular bisector of any chord. It's how you find a circle you can't yet see.
The perpendicular bisector of two points $A$ and $B$ is the line through their midpoint, at right angles to $AB$ — built from two skills you already have: the midpoint formula and the negative-reciprocal rule. What makes it worth a lesson of its own is where else it shows up: a chord of a circle is just a segment joining two points on that circle, and the perpendicular bisector of any chord always passes through the centre.
The centre of a circle is, by definition, the same distance — the radius — from every point on the circle. So the centre is equidistant from the two ends of any chord $A$ and $B$. But the set of all points equidistant from $A$ and $B$ is exactly the perpendicular bisector of $AB$. The centre is one of those points, so it must lie on that line. This is true for every chord you could draw — which means two different chords' bisectors cross at exactly one place: the centre.
If you're given two points on a circle plus one extra condition on the centre (for instance, that it lies on a given line), find the bisector of those two points and intersect it with the condition — one bisector is enough. If you're given three points on a circle and nothing else, you need to intersect two different bisectors (from two different pairs of points) — their crossing point is the centre, and a third bisector would simply pass through the same point as a check.
Worked example 1
If a question just says "the perpendicular bisector of $AB$" with no circle mentioned, this is all there is to it — midpoint plus perpendicular gradient. The circle context in the next two examples adds a use for the line, not a new way to find it.
Worked example 2
Two points alone never pin down a unique circle — infinitely many circles pass through $A$ and $B$, with centres anywhere along the bisector. It always takes a second condition (a given line, a given radius, or a third point) to fix the centre exactly.
Worked example 3
With three points and no other condition given, one bisector isn't enough — find the bisectors of two different chords and intersect them.
Only two bisectors are needed — a third (from bisecting $PR$) would cross at the very same point, since all three lie on the same circle. Checking the distance from $(4,6.6)$ to any two of $P$, $Q$, $R$ gives the same value ($3.4$ to both $P$ and $R$), which confirms the centre is correct without any extra bisector.
Finding the midpoint of a chord and stopping there — the midpoint is on the chord, not the circle's centre. The centre only appears once you find where the perpendicular bisector meets a second condition (another bisector, or a given line). Also watch the case of a vertical or horizontal chord: its perpendicular bisector is horizontal or vertical respectively, with no negative-reciprocal calculation needed at all.
Check yourself
A circle passes through $A(2, 1)$ and $B(2, 9)$, and its centre lies on the line $y = x + 1$. Find the coordinates of the centre.
(Hint: AB is vertical — what does that make its perpendicular bisector?)
A vertical chord flips the usual gradient calculation entirely — no negative reciprocal is needed, since perpendicular-to-vertical is simply horizontal.