Chords and the Perpendicular from the Centre

A circle is rich with named parts: centre, radius, chord, arc, diameter, segment, sector, secant, tangent. This lesson defines them, then proves the most fundamental chord-centre relationship: the perpendicular dropped from the centre to a chord bisects the chord. Together with its converse, this single theorem is used in every chord-related problem.

Definitions

A circle is the set of all points in a plane that are equidistant from a fixed point. That fixed point is the centre; the fixed distance is the radius.

• A chord is a line segment with both endpoints on the circle.
• A diameter is a chord passing through the centre. It is the longest possible chord, with length $2r$ where $r$ is the radius.
• An arc is a portion of the circle between two points. Every chord divides the circle into two arcs , the minor arc (shorter) and the major arc (longer).
• A segment of a circle is the region between a chord and an arc.
• A sector is the region between two radii and an arc.
• A secant is a line that intersects the circle in two points; the chord is the portion inside.
• A tangent is a line that touches the circle at exactly one point (you will study tangents in Class X).

The theorem and its proof

Theorem. The perpendicular from the centre of a circle to a chord bisects the chord.

Given. Circle with centre $O$, chord $AB$, and $OM \perp AB$ with $M$ on $AB$.

To prove. $AM = MB$.

Construction. Join $OA$ and $OB$.

Proof.

Q.E.D.

The proof uses the RHS criterion (right angle, hypotenuse, common side). Elegant.

The converse

Theorem (Converse). The line joining the centre to the midpoint of a chord is perpendicular to the chord.

Given. Circle with centre $O$, chord $AB$, $M$ the midpoint of $AB$.

To prove. $OM \perp AB$.

Proof. Join $OA, OB$. We have $OA = OB$ (radii), $AM = MB$ (midpoint), and $OM = OM$ (common). By SSS, $\triangle OMA \cong \triangle OMB$. Hence $\angle OMA = \angle OMB$ by CPCT. Since they form a linear pair summing to $180^\circ$, each is $90^\circ$. So $OM \perp AB$.

Q.E.D.

Three useful consequences

Consequence 1. Given any chord, the foot of the perpendicular from the centre is the midpoint. Use this to find the midpoint of a chord without measuring.

Consequence 2. Three non-collinear points determine a unique circle. The perpendicular bisectors of any two chords meet at the centre. We will use this in constructions and inscribed-figure problems.

Consequence 3. Equal chords are equidistant from the centre. We prove this in the next lesson.

Worked examples

Example 1. A chord of length $16$ cm is in a circle of radius $10$ cm. Find the distance from the centre to the chord.

The perpendicular bisects the chord into two halves of $8$ cm each. By Pythagoras in the right triangle formed (radius hypotenuse, half-chord one leg): $\sqrt{10^2 - 8^2} = \sqrt{36} = 6$. Distance: $6$ cm.

Example 2. A chord of a circle of radius $13$ is at distance $5$ from the centre. Find the chord length.

Half-chord $= \sqrt{13^2 - 5^2} = \sqrt{144} = 12$. Chord length: $24$.

Example 3. In a circle of radius $r$, the perpendicular from the centre to a chord has length $d$. Express the chord length in terms of $r$ and $d$.

By Pythagoras: half-chord $= \sqrt{r^2 - d^2}$. Full chord $= 2\sqrt{r^2 - d^2}$.

Example 4. Two chords $AB$ and $CD$ of a circle have midpoints $M$ and $N$. Prove that $OM \perp AB$ and $ON \perp CD$.

By the converse, the line from the centre to a chord's midpoint is perpendicular to the chord. Apply to each chord.

Example 5. A diameter of a circle is the longest possible chord. Justify.

A diameter has length $2r$. Any other chord has length less than $2r$ , in fact, equal to $2\sqrt{r^2 - d^2}$ where $d > 0$ is the perpendicular distance from the centre. Since $d > 0$, this is strictly less than $2r$.

Try it yourself

1. State the perpendicular-from-centre theorem and its converse.
2. A chord of length $24$ cm is in a circle of radius $13$ cm. Find the distance from the centre.
3. A chord is at distance $6$ from the centre of a circle of radius $10$. Find the chord length.
4. Prove that the perpendicular from the centre bisects the chord using RHS.
5. The diameter is the longest chord , prove it.
6. In a circle, two chords of length $10$ cm each are drawn. Without further data, can we say their distances from the centre are equal?
7. A chord of length $30$ cm is at distance $8$ from the centre. Find the radius.
8. Define: chord, secant, tangent.
9. Three non-collinear points lie on a unique circle. Where is the centre?
10. A chord subtends $\angle AOB$ at the centre. The perpendicular from $O$ to the chord bisects this angle. Justify.

Pitfalls / Insight

• The perpendicular must come from the centre. A perpendicular from another point may not bisect the chord.
• Both the theorem and its converse are useful. Sometimes you're given a perpendicular, sometimes a midpoint.
• The half-chord and the perpendicular form a right triangle with the radius as hypotenuse. Pythagoras solves length problems immediately.

Insight. A circle is profoundly symmetric: reflection across any diameter is a symmetry. The perpendicular-from-centre theorem is the algebraic form of that symmetry , it forces equal halves on either side of the perpendicular.