Symmetry of the Circle

In the Chinese book: 圆的对称性

Fold a circle on any diameter and it fits itself. That fold proves the theorem.

Try this

Drag the slider to $100%$ . After the red half of the circle flips over, which point does $A$ land on?
Go back to $0%$ and drag $A$ or $B$ . The purple diameter $C D$ turns with them. When are the readouts $A M$ and $M B$ different?
Look at “Arcs $A C$ and $B C$ ”. Besides the chord, what else does the diameter bisect?
Move the chord toward the center (drag both $A$ and $B$ downward). As the distance $O M$ from the center shrinks, does the chord get longer or shorter? How long can it get?

A circle has two kinds of symmetry:

Reflection symmetry: the line through any diameter is an axis of symmetry, so there are infinitely many axes.
Rotational symmetry: turn it about its center by any angle and it lands on itself. So equal central angles stand on equal arcs and equal chords.

Apply the first kind to a chord and you get the perpendicular-diameter theorem:

diameter C D ⊥ A B ⟹ A M = M B, A C ⌢ = B C ⌢, A D ⌢ = B D ⌢

Predict first

A circle has radius $5$ , and a chord is at distance $3$ from the center. How long is the chord?

Choose an answer first; then you can see the result.

Why

Proof: a diameter perpendicular to a chord bisects it

Step 1 / 6

Given and to prove
$C D$ is a diameter and $C D ⊥ A B$ , with foot $M$ . We want to prove $A M = M B$ .

$C D ⊥ A B \Rightarrow A M = M B ?$
Draw two radii
Altitude = median
Another view: fold it
The arcs match too
A free tool for calculations

If the diameter is not perpendicular to the chord

Break it

Counterexample lab: remove one condition

Conclusion: $C D$ bisects chord $A B$

While they are perpendicular, all four lights show ✓ however you drag $A$ and $B$ .

Practice

Exercise 1Basic

A circle has radius $10$ , and chord $A B$ is at distance $6$ from the center. Find the length of $A B$ .

Exercise 2Basic

A chord of a circle has length $24$ and is at distance $5$ from the center. Find the radius.

Exercise 3Advanced

A circle has radius $5$ . Chords $A B = 6$ and $C D = 8$ are parallel. What can the distance between the two chords be? (List every possibility, separated by commas.)

Exercise 4Challenge

Circle $O$ has radius $13$ , and point $P$ is inside it with $O P = 5$ . How many chords through $P$ have integer length?

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Using “a diameter that bisects a chord is perpendicular to it” with no condition

A B

and

C D

are both diameters of circle

O

. Since

C D

passes through the midpoint

O

A B

C D ⊥ A B

What went wrong? Think first, then open

Diagnosis Any two diameters bisect each other at the center, yet they can meet at any angle. The converse of the perpendicular-diameter theorem needs the bisected chord to be not a diameter.

The correct statement: a diameter that bisects a chord (not a diameter) is perpendicular to that chord and also bisects the two arcs it cuts off. Before using it, check that the chord does not pass through the center.

“The converse hides a condition in brackets — the chord is not a diameter”

Exhibit

Two parallel chords, only one picture drawn

In a circle of radius

5

, chords

A B = 6

and

C D = 8

with

A B ∥ C D

. Their distances from the center are

4

and

3

, so the chords are

4 - 3 = 1

apart.

What went wrong? Think first, then open

Diagnosis The problem never says the two chords are on the same side of the center. One can be above it and one below, with the center between them.

Same side of the center:

4 - 3 = 1

. Opposite sides:

4 + 3 = 7

. The answer is

1

7

“When a circle problem comes without a figure, first ask “could it be drawn another way?””