Equal Chords and Their Distances from the Centre

This lesson collects a pair of theorems linking the length of a chord to its distance from the centre. They are converse to each other and together let you reason fluently about chords without measurement: just compare lengths or distances.

The two theorems

Theorem 1. Equal chords of a circle are equidistant from the centre.

Theorem 2 (Converse). Chords of a circle that are equidistant from the centre are equal in length.

Proof of Theorem 1

Given. Circle with centre $O$, two chords $AB$ and $CD$ with $AB = CD$. Let $OM \perp AB$ at $M$ and $ON \perp CD$ at $N$.

To prove. $OM = ON$.

Proof. By the perpendicular-from-centre theorem, $M$ is the midpoint of $AB$ and $N$ is the midpoint of $CD$.

So $AM = \tfrac{1}{2} AB$ and $CN = \tfrac{1}{2} CD$. Since $AB = CD$, we have $AM = CN$.

Consider $\triangle OMA$ and $\triangle ONC$. $OA = OC$ (radii), $AM = CN$ (just shown), and $\angle OMA = \angle ONC = 90^\circ$. By RHS, $\triangle OMA \cong \triangle ONC$. So $OM = ON$ by CPCT. Q.E.D.

Proof of Theorem 2

Given. Circle with centre $O$, two chords $AB$ and $CD$ with perpendicular distances $OM = ON$ from $O$.

To prove. $AB = CD$.

Proof. Consider $\triangle OMA$ and $\triangle ONC$. $OA = OC$ (radii), $OM = ON$ (given), $\angle OMA = \angle ONC = 90^\circ$. By RHS, $\triangle OMA \cong \triangle ONC$. So $AM = CN$ by CPCT. Doubling: $AB = 2 AM = 2 CN = CD$. Q.E.D.

Symmetric statements

The two theorems together say: for chords of the same circle, equality of chord length is equivalent to equality of distance from the centre. Either way, equality of one forces equality of the other.

Applications

Application 1. Two equal chords subtend equal arcs. This is one half of what we will discuss in the next lesson , equal chord $\Rightarrow$ equal angle at the centre $\Rightarrow$ equal arc.

Application 2. In a circle of radius $r$, the locus of midpoints of all chords of length $\ell$ is a circle concentric with the given circle. The midpoints are all at distance $\sqrt{r^2 - (\ell/2)^2}$ from the centre , same distance, so concentric circle.

Application 3. Finding equal chords: in problem-solving, if you know two chords have equal length, you can immediately conclude their perpendicular distances from the centre are equal. This often closes a proof.

Worked examples

Example 1. In a circle of radius $10$, two chords of length $12$ each are drawn. Find the distance of each chord from the centre.

By Pythagoras, perpendicular distance $= \sqrt{10^2 - 6^2} = \sqrt{64} = 8$. Both chords are at distance $8$ from the centre , confirming Theorem 1.

Example 2. In a circle of radius $13$, two chords are at distance $5$ from the centre. Find the length of each.

By Pythagoras, half-chord $= \sqrt{13^2 - 5^2} = 12$. Length $= 24$. Both chords are length $24$ , confirming Theorem 2.

Example 3. Two chords of a circle, lying on opposite sides of the centre, are equal in length. Show that the perpendicular distances from the centre are equal.

By Theorem 1, equal chords are equidistant from the centre. The "opposite sides" detail just specifies geometry; the theorem still applies.

Example 4. A chord $AB$ and a chord $CD$ of a circle have $AB = 10, CD = 10$. The perpendicular from the centre to $AB$ has length $4$. Find the perpendicular to $CD$.

By Theorem 1, equal chords are equidistant: the perpendicular to $CD$ has length $4$.

Example 5. Two chords of a circle of radius $5$ have perpendicular distances $3$ and $4$ from the centre. Find the lengths.

Chord 1: $2\sqrt{25 - 9} = 2 \cdot 4 = 8$. Chord 2: $2\sqrt{25 - 16} = 2 \cdot 3 = 6$. They are different lengths because their distances differ.

Try it yourself

1. State Theorem 1 and prove it using RHS.
2. State Theorem 2 and prove it using RHS.
3. Two chords of a circle of radius $6$ have perpendicular distance $2$ and $5$ from the centre. Find their lengths.
4. A chord of length $20$ is at distance $6$ from the centre. Find the radius.
5. In a circle of radius $r$, the longest chord is $\ldots$ (fill in).
6. Two chords are equal in length. Are they always parallel? Why or why not?
7. The midpoints of all chords of length $\ell$ in a circle lie on which curve?
8. A circle has two chords of length $7$ each, on opposite sides of the centre. Are they parallel?
9. Two chords of length $24$ and $10$ in a circle of radius $13$. Find their distances from the centre.
10. In a circle of radius $5$, find the perpendicular distance from the centre for a chord of length $8$.

Pitfalls / Insight

• Equality goes both ways. Equal chords $\Leftrightarrow$ equidistant from the centre.
• The chord need not pass through the centre. Diameters are the longest chords, but equal-chord theorems work for any chord pair.
• Use Pythagoras as soon as you have the right triangle of radius-perpendicular-half-chord.

Insight. Equal chords and equal-distance relationships are essentially the same statement viewed from two sides. Once you internalise this, you can switch between chord length and distance-from-centre fluidly.