Why the longest chord you can draw always runs through the middle
Try to draw the longest possible chord in a circle and you are forced through the centre, giving the diameter. Slide any chord away from the centre and it shrinks. That single observation is the heart of the rules for equal and unequal chords, and both come from one right triangle.
What Are Equal and Unequal Chords?
A chord is a straight segment joining two points on a circle. Two chords are equal when they have the same length and unequal when they differ. The interesting question is how a chord's length relates to its distance from the centre, measured as the perpendicular distance from the centre to the chord.
Two theorems answer it. Equal chords are always the same distance from the centre, and among unequal chords the longer one is closer. Both rest on one fact about chords of a circle: the perpendicular from the centre to a chord bisects it. That, in turn, is the perpendicular bisector theorem applied inside circles.
By the end you will be able to prove both theorems and compute any chord length from its distance to the centre.
Are Equal Chords Equidistant From the Centre?
Yes. Equal chords of a circle are equidistant from the centre. If two chords have the same length, the perpendicular distances from the centre to each chord are equal. The reason is that equal chords create congruent right triangles with the radius, so the third sides, the perpendicular distances, must match.
This is one of the standard circle theorems, taught alongside the fact that the longest chord, the diameter, is the one that passes through the centre. The diameter of a circle is simply the extreme case: distance zero, maximum length.
How Do You Prove Equal Chords Are Equidistant?
The proof drops each chord into a right triangle and pairs the two triangles by congruence.
Given: A circle with centre $O$ and two equal chords $AB = CD$. Perpendiculars $OM \perp AB$ and $ON \perp CD$.
To prove: $OM = ON$.
Proof, step by step.
The perpendicular from the centre bisects the chord, so:
$$AM = \tfrac{1}{2}AB, \qquad CN = \tfrac{1}{2}CD$$
Since $AB = CD$, halving both gives:
$$AM = CN$$
Now compare right triangles $OMA$ and $ONC$:
$$OA = OC \quad (\text{radii of the same circle})$$
$$AM = CN \quad (\text{just shown})$$
$$\angle OMA = \angle ONC = 90°$$
By RHS congruence in triangles, $\triangle OMA \cong \triangle ONC$, so the corresponding sides are equal:
$$OM = ON$$
The two equal chords are the same distance from the centre. The converse is also true and proved the same way: chords equidistant from the centre are equal in length.
For Unequal Chords, Which Is Nearer the Centre?
Of two unequal chords, the longer chord is nearer the centre. As a chord moves closer to the centre it grows; as it moves away it shrinks, reaching zero when it slips off the circle entirely. The diameter, passing right through the centre, is the longest chord of all.
The clean way to see this is the chord-length formula. For a chord at perpendicular distance $d$ from the centre of a circle of radius $r$:
$$\text{chord length} = 2\sqrt{r^2 - d^2}$$
Here $r$ is the radius and $d$ is the perpendicular distance from the centre to the chord. As $d$ grows, $r^2 - d^2$ shrinks, so the chord shortens; when $d = 0$ the chord is $2r$, the diameter. This inverse link between distance and length is exactly why the longer chord sits nearer the centre. The relationship between the diameter and every other chord is explored further in chords and diameters.
Examples of Equal and Unequal Chords
Example 1
A circle of radius $13$ cm has a chord $10$ cm from the centre. Find the chord's length.
Use the chord-length formula with $r = 13$, $d = 10$:
$$\text{length} = 2\sqrt{r^2 - d^2} = 2\sqrt{13^2 - 10^2} = 2\sqrt{169 - 100} = 2\sqrt{69}$$
The chord is $2\sqrt{69} \approx 16.6$ cm long.
Example 2
Two chords of a circle are $8$ cm and $6$ cm. A student says the $6$ cm chord is nearer the centre because it is smaller. Correct the reasoning.
The instinct that "smaller means closer" runs the rule backwards. Test it against the chord-length formula: a chord grows as it nears the centre, because $r^2 - d^2$ increases when $d$ shrinks. So the larger chord is the closer one.
The $8$ cm chord is the longer, so it is nearer the centre; the $6$ cm chord lies farther out. Longer chord, smaller distance, every time.
Example 3
In a circle of radius $5$ cm, a chord is $6$ cm long. Find its distance from the centre.
Rearrange the chord-length formula. Half the chord is $3$ cm, and it forms a right triangle with the radius and the distance $d$:
$$r^2 = d^2 + \left(\tfrac{\text{chord}}{2}\right)^2$$
$$5^2 = d^2 + 3^2 ;\Rightarrow; d^2 = 25 - 9 = 16 ;\Rightarrow; d = 4 \text{ cm}$$
The chord is $4$ cm from the centre.
Example 4
Two equal chords of a circle are each $16$ cm and one is $6$ cm from the centre. How far is the other from the centre?
By the equal-chords theorem, equal chords are equidistant from the centre, so the second chord is also:
$$d = 6 \text{ cm}$$
No calculation is needed beyond recognising the theorem.
Example 5
A chord $24$ cm long lies in a circle of radius $13$ cm. A second chord in the same circle is $4$ cm from the centre. Which chord is longer?
Find each distance and length. For the $24$ cm chord, half-chord $= 12$:
$$d_1 = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm}$$
The second chord is $4$ cm from the centre, which is less than $5$ cm, so it is nearer the centre and therefore the longer chord. Check by length:
$$\text{length}_2 = 2\sqrt{13^2 - 4^2} = 2\sqrt{169 - 16} = 2\sqrt{153} \approx 24.7 \text{ cm}$$
The second chord ($\approx 24.7$ cm) is longer than the first ($24$ cm), confirming that the nearer chord is longer.
Example 6
A chord passes through the centre of a circle of radius $9$ cm. Find its length using the formula.
A chord through the centre has distance $d = 0$:
$$\text{length} = 2\sqrt{9^2 - 0^2} = 2\sqrt{81} = 18 \text{ cm}$$
The chord is $18$ cm, which is the diameter, $2r$, the longest chord possible. This is the extreme every other chord is measured against, and it separates the two segments of a circle equally.
Where Is the Equal-Chords Property Used?
The equal-chords rule shows up wherever equal-length cuts or supports must sit symmetrically around a centre.
Wheel spokes and flanges. Equal chords equidistant from the hub keep a wheel balanced; unequal spacing shifts the centre of mass and causes wobble.
Arch and bridge design. Equal chords across a circular arch carry load symmetrically, so equidistance from the centre is a stability check.
Lens and mirror grinding. Measuring a chord's length gives its distance from the optical centre through the same $2\sqrt{r^2 - d^2}$ relation, letting a maker verify curvature.
The chord-length relation is a direct consequence of the right triangle formed by radius, half-chord, and centre-distance, a standard result for a chord in geometry. Get the distance wrong and a load-bearing chord lands off-centre. The perpendicular that bisects each chord is examined on its own in the perpendicular bisector of a chord.
Mistakes to Watch For
Mistake 1: Thinking the shorter chord is nearer the centre
Where it slips in: Ranking two unequal chords by distance.
Don't do this: Assume "smaller chord, smaller distance."
The correct way: The rule runs the other way, the longer chord is nearer the centre, because a chord grows as it approaches the centre. The rusher who links "small" with "close" reverses every unequal-chord problem.
Mistake 2: Forgetting to halve the chord in the right triangle
Where it slips in: Using the chord-length formula or its rearrangement.
Don't do this: Put the full chord length as a leg of the right triangle.
The correct way: The perpendicular from the centre bisects the chord, so the leg is half the chord, not the whole. The right triangle uses radius (hypotenuse), distance $d$ (one leg), and half-chord (other leg). The memoriser who plugs in the full chord gets every distance wrong.
Mistake 3: Applying the equal-chords theorem across different circles
Where it slips in: Two circles of different radii.
Don't do this: Claim equal chords are equidistant when the chords belong to circles of different sizes.
The correct way: The theorem holds within one circle, or within congruent circles of the same radius. Chords of the same length in circles of different radii sit at different distances. The second-guesser who ignores the "same circle" condition draws a false conclusion.
Conclusion
Equal chords of a circle are equidistant from the centre, and chords equidistant from the centre are equal.
Of two unequal chords, the longer one lies nearer the centre; the diameter is the longest chord of all.
The chord-length formula $2\sqrt{r^2 - d^2}$ makes the distance-length link exact.
The perpendicular from the centre bisects every chord, which is what powers both proofs.
The most common mistake is thinking the shorter chord is nearer the centre; the rule is the reverse.
To work chord problems through with a teacher, explore Bhanzu's geometry tutor, our middle school math tutor sessions, or math classes online. To watch a trainer prove the equal-chords theorem live, you can book a free demo class.
Read More
Secant of a circle — the line through two points of a circle, the chord extended.
Parts of a circle — centre, radius, chord, and the other named elements.
Inscribed angle theorem — how a chord's angle at the circumference relates to the centre.
Tangents from an external point — the equal-length result for tangents, a cousin of the equal-chords rule.
Common tangents — lines that touch two circles, extending chord-and-tangent geometry.
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