Perpendicular from centre to a chord

This result follows directly after the "equal chords subtend equal angles at the centre" theorem in the NCERT Class 9 Circles chapter, and it is the workhorse for almost every numeric chord problem that follows it.

Result 1 (perpendicular implies bisection). Let $O$ be the centre of a circle and $AB$ a chord. If $OM \perp AB$ with $M$ on $AB$, then $AM = MB$ — the perpendicular from the centre bisects the chord.

Result 2 (converse — bisection implies perpendicular). If $M$ is the midpoint of chord $AB$, then $OM \perp AB$ — the line joining the centre to the midpoint of a chord is perpendicular to that chord. Both directions are prescribed as "state without proof" in the Class 9 syllabus, though the reasoning is short: draw radii $OA$ and $OB$. In right triangles $OMA$ and $OMB$, $OA = OB$ (both radii), $OM$ is common to both triangles, and the angles at $M$ are right angles, so by the RHS congruence rule $\triangle OMA \cong \triangle OMB$, which forces $AM = MB$. The converse follows the same congruence the other way round, using SSS instead of RHS. Because $M$ is a right angle, $\triangle OMA$ is a right triangle with hypotenuse $OA = r$ (the radius), so Pythagoras' theorem gives the working relation used throughout this topic:

where: $r$ is the radius of the circle, $d = OM$ is the perpendicular distance from the centre to the chord, and $AB$ is the length of the chord (so $AB/2$ is the half-chord, $AM$). Given any two of $r$, $d$, and $AB$, this single equation solves for the third — that is the entire toolkit for this section.

The goal is to show, from the two directions of the theorem stated above, that a right triangle formed by the radius, the perpendicular distance from the centre, and half a chord always satisfies $r^2 = d^2 + (AB/2)^2$, so that any exam question giving two of these three quantities can be solved for the third.

Step 1. Set up the triangle. Let $O$ be the centre of a circle of radius $r$, let $AB$ be a chord, and let $M$ be the foot of the perpendicular from $O$ onto $AB$, so $OM \perp AB$ and $OM = d$.

Step 2. Apply Result 1. Since $OM$ is the perpendicular from the centre to the chord, $M$ is the midpoint of $AB$, so $AM = MB = \dfrac{AB}{2}$. This converts "half of an unfamiliar chord" into a labelled side of a triangle you can compute with.

Step 3. Identify the right triangle. In $\triangle OMA$, the angle at $M$ is $90^\circ$ by construction, so $OA$ (a radius, length $r$) is the hypotenuse, while $OM = d$ and $AM = AB/2$ are the two legs.

Step 4. Apply Pythagoras' theorem to $\triangle OMA$: $OA^2 = OM^2 + AM^2$, which is exactly $r^2 = d^2 + \left(\dfrac{AB}{2}\right)^2$. Any one of $r$, $d$, or $AB$ can now be found from the other two by simple substitution and a square root.

Problem. A chord of a circle of radius $13\text{ cm}$ is at a distance of $5\text{ cm}$ from the centre. Find the length of the chord.

Step 1. Set up the triangle. Let $O$ be the centre, $AB$ the chord, and $M$ the foot of the perpendicular from $O$ to $AB$, so $OM = d = 5\text{ cm}$ and $OA = r = 13\text{ cm}$.

Step 2. Since the perpendicular from the centre bisects the chord, $AM = MB = \dfrac{AB}{2}$, and $\triangle OMA$ is right-angled at $M$.

Step 3. Apply Pythagoras: $OA^2 = OM^2 + AM^2$, so $13^2 = 5^2 + AM^2$, giving $169 = 25 + AM^2$, so $AM^2 = 144$ and $AM = 12\text{ cm}$.

Step 4. Since $M$ bisects $AB$, $AB = 2 \times AM = 2 \times 12 = \mathbf{24\text{ cm}}$.

Competitive stretch

Two parallel chords, $AB = 24\text{ cm}$ and $CD = 10\text{ cm}$, lie on opposite sides of the centre $O$ of a circle with radius $13\text{ cm}$. Find the distance between them. Applying the same relation to each chord separately: for $AB$, half-chord $= 12\text{ cm}$, so $13^2 = d_1^2 + 12^2 \Rightarrow d_1 = 5\text{ cm}$; for $CD$, half-chord $= 5\text{ cm}$, so $13^2 = d_2^2 + 5^2 \Rightarrow d_2 = 12\text{ cm}$. Since $AB$ and $CD$ lie on opposite sides of $O$, the distance between them is $d_1 + d_2 = 5 + 12 = \mathbf{17\text{ cm}}$. (Had they been on the same side, the distance would instead be $d_2 - d_1 = 7\text{ cm}$.) One perpendicular-from-centre idea, applied twice, unlocks two-chord problems.

• Using the full chord length instead of the half-chord in Pythagoras. The right triangle's leg is half the chord, not the whole chord — forgetting to halve $AB$ is the single most common slip.
• Assuming any line from the centre to a chord bisects it. Only the perpendicular from the centre is guaranteed to pass through the midpoint; a line at any other angle generally misses it.
• Mixing up the hypotenuse. In $\triangle OMA$, the radius $OA$ is always the hypotenuse, opposite the right angle at $M$ — do not substitute $OM$ or $AM$ in its place.
• Treating a diameter as an exception. If the chord is a diameter, the perpendicular distance from the centre is simply $0$, since the centre lies on the diameter itself — this is a valid, trivial case of the same formula, not a separate rule.

Ask the tutor: want to see this combined with the "equal chords are equidistant from the centre" theorem in a mixed problem — ask the AI tutor to walk you through it.