How a Broken Plate Gives Away Its Missing Center
Hand an archaeologist a curved shard of a smashed pottery plate and they can tell you the diameter of the whole plate, even though the center vanished with the rest of it centuries ago. The trick is a single, quiet theorem: draw any chord across the surviving arc, build its perpendicular bisector, and that line runs dead through the center. Do it for a second chord and the two lines cross exactly where the center used to be.
What Is the Perpendicular Bisector of a Chord?
The perpendicular bisector of a chord is the line that meets the chord at a $90°$ angle and passes through its midpoint. "Perpendicular" gives the right angle; "bisector" gives the equal-halves cut. Do both to a chord of a circle and something special is forced: the line must pass through the center.
Two supporting terms first, so nothing is used cold. A chord is a straight segment joining two points on a circle. A perpendicular bisector of any segment is the line at right angles to it through its midpoint, every point on that line is equidistant from the segment's two endpoints, which is exactly the perpendicular bisector theorem at work.
The circle theorem states it three equivalent ways:
The perpendicular bisector of a chord passes through the center.
The perpendicular from the center to a chord bisects the chord.
The line from the center to a chord's midpoint is perpendicular to the chord.
All three describe the same right angle at the same midpoint, read from different starting facts.
Why Does the Perpendicular Bisector of a Chord Pass Through the Center?
The center of a circle is the same distance, the radius, from every point on the circle. So the center is equidistant from the chord's two endpoints $A$ and $B$, because both are radii.
Now recall what the perpendicular bisector of $AB$ actually is: the set of all points equidistant from $A$ and $B$. The center is equidistant from $A$ and $B$, so the center must lie on that set, on the perpendicular bisector. There is no wiggle room; the theorem falls straight out of "all radii are equal."
How Do You Prove the Theorem?
The formal proof uses triangle congruence. Let $O$ be the center, $AB$ a chord, and $M$ its midpoint. Join $OA$, $OB$, and $OM$.
$$OA = OB \quad (\text{both radii})$$
$$AM = BM \quad (M \text{ is the midpoint of } AB)$$
$$OM = OM \quad (\text{common side})$$
By the SSS criterion, $\triangle OAM \cong \triangle OBM$. Congruent triangles have equal corresponding angles, so:
$$\angle OMA = \angle OMB$$
But $\angle OMA$ and $\angle OMB$ sit on a straight line at $M$, so they add to $180°$. Two equal angles summing to $180°$ are each $90°$:
$$\angle OMA = \angle OMB = 90°$$
So $OM$ is perpendicular to $AB$ and passes through its midpoint, $OM$ is the perpendicular bisector, and it contains the center $O$. This is the reasoning descended from Euclid's Elements, whose author is profiled at MacTutor.
The converse is equally useful: if a line from the center meets a chord at a right angle, then it bisects the chord. Same congruent triangles, run the other direction.
How Do You Find the Center of a Circle Using Chords?
This is the theorem's most-used application, and the answer to the pottery-shard puzzle.
Draw the first chord anywhere across the circle and construct its perpendicular bisector. The center lies somewhere on this line.
Draw a second, non-parallel chord and construct its perpendicular bisector. The center lies on this line too.
Take the intersection. The only point on both perpendicular bisectors is the center, because every chord's perpendicular bisector passes through it.
The two chords must be non-parallel. Parallel chords give parallel perpendicular bisectors, which never meet, so they can never pin down a single point.
What Are the Properties Around Chord Bisectors?
The theorem sits inside a small family of chord-and-center facts worth carrying together.
The center lies on every chord's perpendicular bisector. Pick any chord; its perpendicular bisector runs through the center.
A perpendicular from the center bisects the chord (the converse), and the segment $OM$ is the chord's distance from the center.
Equal chords are equidistant from the center, so their bisecting perpendiculars have equal length inside the circle.
Only two non-parallel chords are needed to locate the center, a third is a redundant check.
The perpendicular bisector of a diameter is another diameter (through the same center), since the center sits on it and it is a chord too, tying back to chords and diameters.
Examples of Perpendicular Bisector of a Chord
Example 1
A chord $AB$ has midpoint $M$, and a line through the center $O$ meets $AB$ at $M$. What angle does $OM$ make with $AB$?
By the converse of the theorem, a line from the center through the midpoint of a chord is perpendicular to it.
$$\angle OMA = 90°$$
Final answer: $OM$ is perpendicular to $AB$; the angle is $90°$.
Example 2
You want to find the center of a circle drawn on paper. You carefully construct the perpendicular bisector of one chord and stop there. Have you found the center?
Wrong attempt. One perpendicular bisector "passes through the center," so it feels like the job is done, pick any point on the line and call it the center.
Break. But a line is infinitely many points. The center is just one of them. A single bisector narrows the center to a line, not a point, so you cannot yet say where it is.
Correct. Construct the perpendicular bisector of a second, non-parallel chord. The center is the unique intersection of the two bisectors.
Final answer: no, you need two non-parallel chords; their bisectors cross at the center.
Example 3
A chord $12$ cm long has its midpoint $8$ cm from the center. Find the radius.
The perpendicular from the center bisects the chord, so half the chord is $6$ cm, and it forms a right triangle with the $8$ cm distance and the radius:
$$r = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}$$
Final answer: the radius is $10$ cm.
Example 4
Two students each construct the perpendicular bisector of a different chord of the same circle. Where do their two lines meet?
Every chord's perpendicular bisector passes through the center, so both lines contain the center. Their intersection is the center of the circle.
Final answer: the two bisectors meet at the center.
Example 5
A perpendicular from the center meets a chord at point $M$, and $AM = 5$ cm. How long is the whole chord $AB$?
The perpendicular from the center bisects the chord, so $M$ is the midpoint and $BM = AM = 5$ cm.
$$AB = AM + BM = 5 + 5 = 10 \text{ cm}$$
Final answer: the chord is $10$ cm long.
Example 6
Why must two chords used to find the center be non-parallel?
Parallel chords have perpendicular bisectors that are also parallel to each other. Parallel lines never intersect, so they cannot single out one point, the center stays undetermined.
Final answer: parallel chords give parallel bisectors that never meet, so no single center point emerges.
Where Is This Used?
Any time someone needs the center of a circle they cannot see whole, this theorem does the work.
Restoration and archaeology. The radius of a broken wheel, plate, or arch is recovered from two chords on the surviving piece.
Machining and manufacturing. Finding the exact center of a circular part (to drill a hole through it) uses two chords and their bisectors.
Surveying circular boundaries. Locating the center of a circular field or roundabout from points on its edge.
Computer graphics. Fitting a circle through sampled points reduces to intersecting perpendicular bisectors, the same idea, run by an algorithm.
The theorem turns "I only have part of the circle" into "I can find the whole thing," which is why it outlives any single problem set.
Circle Chord Theorems - Where Things Go Sideways
Mistake 1: Dropping the "perpendicular" or the "bisector" half
Where it slips in: applying the theorem when a line only bisects the chord, or only meets it at a right angle, but not both.
Don't do this: assuming any line through a chord's midpoint passes through the center.
The correct way: the line must be both perpendicular and a bisector to be guaranteed through the center. A midpoint line that is slanted, or a perpendicular that misses the midpoint, gives you nothing. Both conditions together are what force the center.
Mistake 2: Using parallel chords to find the center
Where it slips in: locating the center from two chords that happen to be parallel.
Don't do this: constructing two perpendicular bisectors and expecting them to cross.
The correct way: choose non-parallel chords. The memorizer who recalls "two chords find the center" but forgets the non-parallel condition ends up with two parallel bisectors that never meet, and no center. The rule only works when the chords head in different directions.
Mistake 3: Reporting the half-chord as the distance from the center
Where it slips in: right-triangle problems mixing up which leg is which.
Don't do this: treating the perpendicular distance $OM$ and the half-chord $AM$ as interchangeable.
The correct way: $OM$ is the perpendicular distance from the center to the chord; $AM$ is half the chord. They are the two legs of the right triangle with the radius as hypotenuse, different roles, different lengths. Label them before plugging into Pythagoras.
Conclusion
The perpendicular bisector of a chord always passes through the center of the circle.
It follows from a single fact: the center is equidistant from the chord's two endpoints, so it lies on their perpendicular bisector.
The formal proof uses SSS congruence of the two triangles the bisector creates.
The converse, a perpendicular from the center bisects the chord, is equally usable.
Two non-parallel chords and their perpendicular bisectors locate the center of any circle.
To take the perpendicular bisector of a chord further with a teacher, explore Bhanzu's geometry tutor, a middle school math tutor, or live math classes online.
Practice These to Solidify Your Understanding
Work through these three: (1) a chord $10$ cm long has its midpoint $12$ cm from the center, find the radius; (2) construct the center of a circle drawn on paper using two chords; (3) a perpendicular from the center meets a chord at $M$ with $AM = 7$ cm, find the full chord length. If the two-chord construction is unclear, return to How Do You Find the Center of a Circle Using Chords? above. Want a live Bhanzu trainer to walk through more circle theorems? Book a free demo class.
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