Equal chords and their distances

Two chords of a circle can sit anywhere on the boundary — one near the top, one near the bottom — yet if they happen to be the same length, they are always tucked away at exactly the same "depth" from the centre. This is the third big result in the Class 9 Circles chapter, following equal chords subtending equal angles at the centre and the perpendicular from the centre bisecting a chord. Here the idea is a two-way link: equal chords are equidistant from the centre, and chords that are equidistant from the centre are equal. Both directions are stated without proof at this level, but understanding why they feel true — through the right-triangle relation connecting radius, distance and half-chord — makes the result far easier to use confidently, and to combine with the earlier theorems in multi-step problems.

The core insight

In a circle, or in congruent circles, chords are equal in length exactly when they lie at equal perpendicular distances from the centre.

The core idea

The distance of a chord from the centre is the length of the perpendicular dropped from the centre onto that chord. If $O$ is the centre and $OM \perp AB$ , then the distance of chord $AB$ from the centre is simply the segment $OM$ — the same quantity that appeared as $d$ in the perpendicular-bisector-of-a-chord result. The theorem links two chords $AB$ and $CD$ of a circle with centre $O$ , where $OM \perp AB$ and $ON \perp CD$ :

AB = CD \iff OM = ON

where: $AB$ and $CD$ are the two chord lengths, and $OM$ , $ON$ are their perpendicular distances from the centre $O$ . The double arrow means the statement works in both directions — equal chords force equal distances, and equal distances force equal chords. The restriction to matters: this holds for chords within one circle, or for chords of congruent circles (same radius). It says nothing about chords of the same length lying in circles of different sizes.

Why it feels true. Recall the relation connecting radius, distance from centre and half the chord, from the perpendicular-bisector topic:

r^2 = d^2 + \left(\frac{\text{chord}}{2}\right)^2

where: $r$ is the radius of the circle, $d$ is the perpendicular distance of the chord from the centre, and chord is its full length. Since $r$ is fixed once the circle is fixed, this equation ties $d$ and the chord length together directly — pick one, and the other is determined. A smaller $d$ must go with a larger chord, and an equal $d$ can only go with an equal chord. This is exactly why the diameter (where $d = 0$ ) is the longest possible chord, and why chords shrink steadily as they move farther from the centre.

Figure

Two equal chords at equal distances from the centre of a circle

Derivation ADistance of a chord from the length of the chord and the radius

Given a circle of radius $r$ and a chord of known length, find the perpendicular distance $d$ of that chord from the centre, and show that this same computation forces two chords of equal length in the same circle to sit at equal distances from the centre — the working link between the equal-chords theorem and the earlier radius-distance-half-chord relation.

Step 1. Drop the perpendicular. Let $M$ be the foot of the perpendicular from centre $O$ to a chord $XY$ , so $OM = d$ is the distance we want, and $OX = r$ is a radius drawn to one end of the chord. Since the perpendicular from the centre to a chord bisects it, $XM = \frac{1}{2}XY$ .

Step 2. Apply Pythagoras in right triangle $OMX$ , which is right-angled at $M$ : $OX^2 = OM^2 + XM^2$ , that is $r^2 = d^2 + \left(\frac{XY}{2}\right)^2$ . Rearranging for the distance gives $d = \sqrt{r^2 - \left(\frac{XY}{2}\right)^2}$ .

Step 3. See why equal chords force equal distances. If a second chord $PQ$ in the same circle has $PQ = XY$ , then it has the identical half-chord length $\frac{PQ}{2} = \frac{XY}{2}$ , and the same radius $r$ appears in its own right triangle. Substituting into the same formula gives the identical value of $d$ . There is no room for the two distances to differ once the chord lengths and the radius match — this is the algebraic reason the theorem holds, even though a formal proof is not required at this level.

Step 4. Read the converse the same way. If instead two chords are given as equidistant ( $d$ the same for both), the equation $r^2 = d^2 + \left(\text{chord}/2\right)^2$ forces $\left(\text{chord}/2\right)^2$ to be the same fixed value for both, so both chords come out equal in length.

Worked example

Problem. In a circle with centre $O$ , chords $AB = 10\text{ cm}$ and $CD = 10\text{ cm}$ . If the distance of $AB$ from $O$ is $3.4\text{ cm}$ , find the distance of $CD$ from $O$ .

Step 1. Note that $AB$ and $CD$ are chords of the same circle and $AB = CD = 10\text{ cm}$ .

Step 2. Apply the theorem directly: equal chords of a circle are equidistant from the centre, so the distance of $CD$ from $O$ equals the distance of $AB$ from $O$ , giving $3.4\text{ cm}$ with no further calculation needed. The distance of $CD$ from $O$ is $3.4\text{ cm}$ .

Competitive stretch

Problem. Two chords $PQ$ and $RS$ of a circle with centre $O$ and radius $17\text{ cm}$ are equidistant from the centre, each at a distance of $8\text{ cm}$ . Find the length of each chord.

Step 1. Let $M$ be the foot of the perpendicular from $O$ to $PQ$ , so $OM = 8\text{ cm}$ and $OP = r = 17\text{ cm}$ . Since $PQ$ and $RS$ are equidistant, the converse of the theorem already guarantees $PQ = RS$ — but the actual length still needs Pythagoras.

Step 2. In right triangle $OMP$ : $OP^2 = OM^2 + MP^2$ , so $17^2 = 8^2 + MP^2$ , giving $289 = 64 + MP^2$ and $MP^2 = 225$ , so $MP = 15\text{ cm}$ .

Step 3. Since $M$ bisects $PQ$ : $PQ = 2 \times MP = 30\text{ cm}$ . The same computation applies unchanged to $RS$ , since it sits at the identical distance $8\text{ cm}$ in the same circle of radius $17\text{ cm}$ . $PQ = RS = 30\text{ cm}$ , matching what the converse predicted directly.

Common traps

Applying the theorem across circles of different radii. "Equal chords are equidistant" holds within one circle, or between congruent circles (equal radii) — not for equal-length chords in circles of different sizes.
Confusing distance from the centre with chord length. These move oppositely: chords closer to the centre are longer, not shorter, up to the diameter (the longest chord, at $d = 0$ ).
Ignoring that "distance" means perpendicular distance. Measuring along any slanted line to the chord does not give the distance used in this theorem — only the perpendicular foot counts.
Forgetting Pythagoras when an actual length or distance is asked for. The equidistant-chords theorem alone proves the chords are equal; a numeric answer still needs $r^2 = d^2 + (\text{chord}/2)^2$ .

Ask the tutor: Want to work through a problem that combines this theorem with the perpendicular-bisector-of-a-chord result in one multi-step question? Ask the AI tutor about "equal chords and their distances."

Why does this theorem only apply to one circle or congruent circles, and not to any two circles

The proof works through $r^2 = d^2 + (\text{chord}/2)^2$ , where $r$ is fixed for a given circle. If two chords of equal length sit in circles of different radii, this equation gives a different $d$ for each circle, because $r$ itself differs. So equal chord lengths in different-sized circles generally sit at different distances from their respective centres — the theorem only becomes reliable when the radius is shared, either literally (one circle) or by being equal (congruent circles).

Do I need to memorise a proof of this theorem for my exam

No. NCERT states this result, in both directions, "without proof" at the Class 9 level — you are expected to know the statement and apply it confidently to numerical problems, not to reproduce a formal two-column proof. What is worth understanding, though, is why it is true, using the radius-distance-half-chord relation, since that reasoning often helps you solve multi-step problems that combine this theorem with the perpendicular-bisector-of-a-chord result.

If I am told two chords are equal, do I still need to use Pythagoras to find their distances

Only if you are asked for an actual numeric distance and only one of the two distances is given directly — then the theorem alone hands you the other distance immediately, with zero extra computation, as in the first worked example. Pythagoras is needed only when neither distance is given upfront and you must compute an actual length or distance from the radius, as in the competitive stretch example.

Equal chords and their distances

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads