Circle Theorems on Chords and Circumcircle

This page explains three important circle theorems step by step. Before starting, make sure you understand the ideas of centre, radius, chord, perpendicular line and perpendicular bisector.

Quick revision

All radii of the same circle are equal.
Both endpoints of a chord lie on the circle.
A perpendicular bisector divides a line segment into two equal parts and is perpendicular to it.
Corresponding sides and angles of congruent triangles are equal.

⭐ Exactly one circle can be drawn through three non-collinear points

This proof is useful for understanding the idea of a circumcircle. Whether the complete proof is required in an examination depends on your syllabus.

Given:	A, B, C are three fixed non-collinear points.
To prove:	One and only one circle can be drawn through A, B, C.
Construction:	Join AB and BC. Draw the perpendicular bisector PQ of AB and the perpendicular bisector RS of BC. Let them intersect at O. Let PQ meet AB at D and RS meet BC at E. Join OA, OB and OC.
Proof:	In △OAD and △OBD,
	AD = DB [D is the midpoint of AB]
	∠ODA = ∠ODB = 90° [PQ ⟂ AB]
	OD = OD [Common side]
∴	△OAD ≅ △OBD [SAS congruence]
∴	OA = OB [Corresponding sides of congruent triangles]
	Similarly, because O lies on the perpendicular bisector RS of BC, the congruence of △OBE and △OCE gives
∴	OB = OC.
∴	OA = OB = OC.
Thus,	A, B, C are all at the same distance from O.
∴	A circle with centre O and radius OA passes through all three points A, B, C.
∴	A circle can be drawn through three non-collinear points.
	Since A, B, C are fixed, the segments AB and BC are fixed. Therefore their perpendicular bisectors PQ and RS are also fixed.
Also,	Since A, B, C are non-collinear, AB and BC do not lie on the same straight line. Hence their perpendicular bisectors meet at one fixed point O.
∴	The centre O is fixed.
Again,	Since O and A are fixed, the radius OA is also fixed.
	Only one circle can be drawn with a fixed centre and a fixed radius.
∴	Exactly one circle can be drawn through the three non-collinear points A, B, C. [Proved]

What new idea do we get from this theorem?

The point where the perpendicular bisectors of two sides of △ABC intersect is equidistant from all three vertices.

This point is the circumcentre of the triangle, and the circle passing through all three vertices is its circumcircle.

⭐ A perpendicular drawn from the centre of a circle to a chord bisects the chord

Given:	In a circle with centre O, AB is a chord other than a diameter. From O, OD is drawn perpendicular to AB. Thus OD ⟂ AB.
To prove:	AD = DB.
Construction:	Join OA and OB.
Proof:	Since OD ⟂ AB, △OAD and △OBD are right triangles.
Now,	OA = OB [Radii of the same circle]
	OD = OD [Common side]
	∠ODA = ∠ODB = 90°.
∴	△OAD ≅ △OBD [RHS congruence]
∴	AD = DB [Corresponding sides of congruent triangles]
∴	D is the midpoint of chord AB.
Therefore,	A perpendicular drawn from the centre of a circle to a chord bisects the chord. [Proved]

Application of the theorem

If AB = 18 cm and OD ⟂ AB, then

AD = DB = 18 ÷ 2 = 9 cm.

So, once the foot of the perpendicular from the centre to a chord is known, the midpoint of the chord is known immediately.

⭐ If the centre of a circle is joined to the midpoint of a chord, the joining line is perpendicular to the chord

This is the converse of the previous theorem.

Given:	AB is a chord, other than a diameter, of a circle with centre O. D is the midpoint of AB, so AD = DB. Join O to D.
To prove:	OD ⟂ AB.
Construction:	Join OA and OB.
Proof:	In △OAD and △OBD,
	OA = OB [Radii of the same circle]
	AD = DB [Given]
	OD = OD [Common side]
∴	△OAD ≅ △OBD [SSS congruence]
∴	∠ODA = ∠ODB [Corresponding angles of congruent triangles]
Now,	A, D, B are collinear. Therefore ∠ODA and ∠ODB form a linear pair and their sum is 180°.
So,	∠ODA + ∠ODB = 180°.
Also,	∠ODA = ∠ODB.
∴	Each angle is 180° ÷ 2 = 90°.
∴	∠ODA = ∠ODB = 90°.
∴	OD ⟂ AB. [Proved]

⭐ Understand the previous two theorems together

Perpendicular from the centre to a chord	⇒ passes through the midpoint of the chord.
Line from the centre to the midpoint of a chord	⇒ is perpendicular to the chord.

Thus, for a chord other than a diameter, the perpendicular from the centre to the chord and the line joining the centre to the midpoint of the chord are the same line.

⭐ Short solved examples

Example 1: Finding the two parts of a chord

AB is a chord of length 20 cm. From the centre O, OD is drawn perpendicular to AB.

By the theorem,

AD = DB.

Therefore,

AD = DB = 20 ÷ 2 = 10 cm.

Answer: AD = DB = 10 cm.

Example 2: How can we prove that a line is perpendicular to a chord?

D is the midpoint of chord AB, and O is the centre of the circle.

Since

AD = DB,

and OD joins the centre to the midpoint of the chord, the converse theorem gives

OD ⟂ AB.

Answer: OD is perpendicular to chord AB.

Example 3: Radius, distance from centre and chord length

A circle has radius 13 cm. The perpendicular distance from the centre O to chord AB is 5 cm. Find the length of AB.

Let OD ⟂ AB. Then D is the midpoint of AB.

In right triangle △ODA,

OA² = OD² + AD² 13² = 5² + AD² AD² = 169 - 25 = 144

AD = 12 cm.

Therefore,

AB = 2AD = 24 cm.

Answer: 24 cm.

✍️ Practice problems

A chord AB is 16 cm long. The perpendicular from the centre meets it at D. Find AD and DB.
D is the midpoint of chord AB of a circle with centre O. What is the value of ∠ODB?
A circle has radius 10 cm. The distance from its centre to a chord is 6 cm. Find the length of the chord.
Explain why three collinear points cannot determine a finite circle.
The perpendicular bisectors of two sides of △ABC meet at O. If OA = 7 cm, find OB and OC.
A chord AB of a circle is 24 cm long and its distance from the centre is 5 cm. Find the radius of the circle.

Answers

AD = DB = 8 cm
90°
16 cm
The perpendicular bisectors needed to locate a finite common centre do not meet at one finite point for three distinct collinear points.
OB = OC = 7 cm
13 cm

⚠️ Common mistakes and how to avoid them

🔸 Do not forget the words “non-collinear” in the first theorem. Three points on the same straight line cannot lie on one finite circle.

🔸 In the converse theorem, proving only that two angles are equal is not enough. Since they form a linear pair, their sum is 180°. Equal angles with sum 180° must each be 90°.

🔸 Do not confuse RHS and SSS congruence. First identify what information is given in each pair of triangles, then choose the correct criterion.

🔸 The perpendicular-from-centre theorem applies to a chord. In numerical problems, remember that the perpendicular divides the whole chord into two equal halves.

📚 Important points to remember

🔹 Exactly one circle passes through three fixed non-collinear points.

🔹 The centre of that circle can be found by intersecting the perpendicular bisectors of two joining segments.

🔹 The common point of the perpendicular bisectors of the sides of a triangle is its circumcentre.

🔹 A perpendicular drawn from the centre of a circle to a chord bisects the chord.

🔹 A line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.

🔹 The last two results are converses of each other.

Quick revision table

Three non-collinear points	determine one and only one circle.
OD ⟂ AB	⇒ AD = DB.
AD = DB	⇒ OD ⟂ AB [where O is the centre].

Revise the basic concepts of a circle first

If the ideas of chord, diameter, arc, segment or sector are not clear, study Basic Concepts of Circle before continuing with the later circle theorems.