We all know that circle theorems are used for solving different geometry problems. Before we learn about these theorems in detail, let us first understand the meaning of a circle. A circle is the locus of a point which moves in a plane so that its distance from a fixed point in the plane is always constant. The fixed point is called the centre of the circle and the constant distance is called its radius.

Here, C = centre, and CP = constant distance = radius

The region occupied by the circle is known as the area of a circle and the outer line of the circle is known as the circumference of the circle. The line which is perpendicular to the circle at any point on the circle is known as the tangent of a circle.

In this mathematics article, we are going to learn about circles and their applications, the theorem of circles and their proof in detail. Uses the circle theorem in real life and also solves some problems based on the circle theorem to understand the topic easily.

Circle Theorem

Circle theorems are special rules in geometry that explain how angles, lines, and parts of a circle relate to each other. These theorems help us understand important facts about circles, such as how tangents, chords, radii, and diameters behave. In simple terms, they show how angles inside or around a circle are connected.

Using circle theorems, we can figure out unknown angles and solve problems involving circles. When we draw shapes, lines, and angles inside a circle, we often notice patterns. These patterns form the basis of circle theorems, which are useful in both math problems and real-life situations like engineering and design.

Circle Theorem Terms

Now we are going to study the different parts of the circle which are as follows:

• Centre of a circle: The centre of a circle is a fixed point that is equidistant from all the points on the boundary of circles.
• Radius of a circle: The radius of a circle is a fixed distance between the centre of the circle and at any point on the boundary of the circle.
• Diameter of a circle: A diameter of a circle is the line segment whose both endpoints are on the circle and it is the largest chord of the circle.
• Chord of a circle: The chord of a circle is the line segment that touches the circle at two different points on the boundary of the circle and divides the circle into two parts.
• Tangent of a circle: The tangent to a circle is a coplanar straight line that touches the circle at a single unique point.
• Circumference of a circle: The circumference of a circle is defined as the linear distance around it. In other words, if a circle is opened to a straight line, then the length of that straight line will be the circumference of a circle.

• Segment of a circle: The segment of a circle is the area contained by the chord and the corresponding arc of a circle. There are two types of segments, i.e., minor segment and major segment. It should be noted that segments do not include the centre.
• Sector of a circle: The sector of a circle is the area circumscribed by two radii and the corresponding arc in a circle. There are two types of sectors, i.e., the minor sector and the major sector.
• Arc of a circle: An arc of a circle is related to a curve, which is a section or portion of its circumference.
• Secant of a circle: A secant is a straight line crossing the circle at two points on the circumference.

Circle Theorems Proof

There are seven main circle theorems:

• Alternate segment circle theorem
• The angle at the centre circle theorem
• Angles in the same segment circle theorem
• Angle in a semicircle theorem
• Chord circle theorem
• Cyclic quadrilateral theorem
• Tangent circle theorem

Below is the statement for each theorem along with their diagram and proof of the theorem.

Theorem 1: Alternate segment theorem

The angle that lies between a tangent and a chord is equal to the angle subtended by the same chord in the alternate segment.

Proof:

Let P be the point on the circumference of the circle and O be the centre of the circle.

AB is the tangent passing through the point P.

The tangent makes \( \angle\alpha \) with the chord PQ.

Consider an \( \angle \) PRQ = \( \beta \) in the alternate segment.

To prove: \( \angle\alpha \) = \( \angle\beta \).

OP = OQ (since both are the radii of the circle).

\( \angle \) OPQ = \(\angle \) OQP (since, the angles opposite to the equal sides are equal).

\( \bigtriangleup \) OPQ is isosceles,

\( \angle \) POQ = \(180^{\circ} \) – \( \angle \) OPQ – \( \angle \) OQP

= \(180^{\circ} \) – 2 \( \angle \) OPQ ……..(i)

Since, AB is the tangent,

\( \angle \) OPB = \(90^{\circ} \) – \( \angle \) OPQ ……..(ii)

From equations (i) and (ii), we find that

\( \angle \) POQ = 2 \( \alpha \).

We know that the angle at the centre is twice the angle at the circumference.

\( \angle \) POQ = 2 \( \angle \) PRQ

\( \angle \) PRQ = \( \frac{1}{2}\angle \) POQ

\( \angle\beta \) = \( \frac{1}{2}(2\alpha) \)

= \( \angle\alpha \).

Thus, the alternate segment theorem is proved.

Learn about Difference Between Circle and Sphere and Symmetry in a Circle

Theorem 2: Angle at the centre theorem

The angle subtended by a chord at the centre is twice the angle subtended by it at the circumference.

Proof:

We have,

• A, B and D are points on the circumference and C is the centre
• BC and CD are radii
• AB and AD are chords

Connect the points AC, we have two triangles ACD and ABC.

In triangle ACD,

AC are CD radii and so this is an isosceles triangle.

Therefore, the angles at A and D are equal (say ‘\(x \)’).

The angles in a triangles add up to \(180^{\circ} \), then the angle at C will be \(180^{\circ} \) – 2 \(x \).

In triangle ABC,

BC and AC are radii and so this is also an isosceles triangle.

Therefore, the angles at A and B are equal (say ‘\(y \)’).

Then the angle at C is \(180^{\circ} \) – 2 \(y \).

The angle at A is \(x\) + \(y\) and the angle at C is sum of \(180^{\circ} \) – 2 \(x\) and \(180^{\circ} \) – 2 \(y\).

We need to know the angle BCD, we can use the fact that angles around a point are \( 360^{\circ} \).

Then angle at BCD = \(360^{\circ} \) – (\(360^{\circ} \) – 2\(x\) – 2\(y\))

= 2\(x\) + 2\(y\).

= 2(\(x\) + \(y\)).

So as the angle BAD = \(x\) + \(y\) and angle at BCD = 2(\(x\) + \(y\)), then the angles at the centre is twice the angle at the circumference.

Theorem 3: Angles in the same segment theorem

Angles in the same segment are equal.

Proof:

Let the two angles at the circumference be ‘a’ and ‘b’.

Plot the centre of a circle and draw two radii from the centre to the circumference (here these are dashed lines). Assume the angle between the two radii to be ‘c’.

Let us inspect angles a and c.

We know that the angle at the centre is twice the angle at the circumference. This means that if the angle at the centre is 2\(x\), the angle at the circumference (i.e. angle a) is now equal to \(x\).

Let us inspect angles b and c.

Again, we know that the angle at the centre is twice the angle at the circumference and so if say that angle c is equal to 2\(x\), angle b is equal to \(x\).

We now have angles ‘a’ and ‘b’ are \(x\) and the angle ‘c’ is equal to 2\(x\).

This means that the two angles at the circumference are the same size and therefore we can state that the angles in the same segment are equal.

Theorem 4: Angles in a semicircle theorem

The angle in a semicircle is 90 degrees.

Proof:

We have,

• Triangle ABC where A, B and C are points on the circumference of a circle.
• AB is the diameter of the circle, passing through the centre.

Connect the centre to point C to create two smaller triangles.

In triangle OAC,

OA and OC are radii and so triangle OAC is isosceles.

Therefore, the two angles at A and C are equal (say ‘\(x\)’).

In triangle OBC,

OB and OC are radii and so this is also an isosceles triangle.

Therefore, the two angles at B and C are equal (say ‘\(y\)’).

Now we have, angle ACB equal to \(x\) + \(y\) which is a sum of two angles \(x\) and \(y\) from the two triangles OAC and OBC.

As the sum of angles in a triangle is equal to \(180^{\circ} \), we have

\(x\) + \(y\) + \(x\) + \(y\) = \(180^{\circ} \)

2\(x\) + 2\(y\) = \(180^{\circ} \)

\(x\) + \(y\) = \(90^{\circ} \)

As \(x\) + \(y\) = \(90^{\circ} \), we can now state that the angle in a semicircle is 90 degrees.

Theorem 5: Chord of a circle theorem

The perpendicular from the centre of a circle to a chord bisects the chord (splits the chord into two equal parts).

Proof:

Consider a circle with a centre at C. The line AB is a chord and CE is a radius. The lines CE and AB intersect at the point D at \(90^{\circ}\) to one another because they are

perpendicular.

Now draw the lines AC and BC. The length AC = BC as they are both radii of the circle.

Triangles ACD and BCD are both right-angle triangles, their hypotenuse is equal and the line CD is the same as it is shared between both triangles. This means that the triangles are congruent and so the line segment BD and the line segment AD are the same lengths or BD = AD.

Hence, the perpendicular from the centre of a circle to a chord bisects the chord into two equal parts.

Theorem 6: Cyclic quadrilateral theorem

The opposite angles in a cyclic quadrilateral are supplementary.

Proof:

Figure ABCD is a quadrilateral inscribed in a circle with centre O.

Draw two radii OB and OD, we have two smaller quadrilaterals.

Let if the angle at A is ‘a’, then the angle at BOD is equal to ‘2a’ (by theorem 2).

Also if the angle at C is ‘c’, then the angle at BOD is equal to ‘2c’ (by theorem 2).

This means we have the point at the centre with the angles ‘2a’ and ‘2c’.

We know angles around a point are equal to \(360^{\circ}\). Therefore, we have

2a + 2c = \(360^{\circ}\)

a + c = \(180^{\circ}\)

Hence, the opposite angles in a cyclic quadrilateral are supplementary.

Theorem 7: Tangent of a circle theorem

Tangents which meet at the same point are equal in length.

Proof:

Firstly we take an arbitrary point A (a random point in space) outside of the circle.

Point A can be connected to the circle by two tangents. One line touches the circle at B, the other tangent touches the circle at C.

Prove The length AB = AC.

If we join OA together and then connect OB and OC, we get two triangles. If we can prove that these two triangles are congruent, then AC will be equal to AB.

• The angles OBA and OCA are \(90^{\circ}\) each as tangents meet a circle at \(90^{\circ}\). This means we have two right-angle triangles.
• As OB and OC are radii of the circle, they must be the same length.

Triangles AOB and AOC are both right angles. They share the same side length AO and another side length of each triangle is the same as it is the radius of the circle (OB = OC). This means that the two triangles are congruent and so AC = AB.

Hence, the tangents that meet at the same point are equal in length.

Learn about Concentric Circles

1. In the given figure, the triangle ABC is inscribed in a circle with centre O. The tangent DE meets the circle at point A. Calculate the size of the angle ABC.

Solution: We have,

• The tangent DE
• The chord AC (that meets the tangent)
• The \(\angle \) CAE = \(56^{\circ} \)
• The \(\angle \) ABC = \(\theta \)

We already knew that the \(\angle \) CAE = \(56^{\circ} \), we do not need to use any other angle fact to determine this angle.

\(\angle \) ABC = \(56^{\circ} \), as the angle in the alternate segment are equal to the angle between the tangent and the associate chord.

2. In the given figure, ABCD is an arrowhead where C is the centre of the circle and A, B, and D lie on the circumference. Calculate the size of the angle BAD.

Solution: We have,

• The \(\angle \) BCD = \(150^{\circ} \)
• BC = CD = Radii
• AB and AD are chords
• The \(\angle \) BAD = \(\theta \)

We already knew that the \(\angle \) BCD = \(150^{\circ} \), we do not need to use any other angle fact to determine this angle.

The angle at the centre is twice the angle at the circumference and so as we know the angle at the centre, we need to divide this number by 2 to get the \(\angle \) BAD.

\(\angle \) BAD = 150 \(\div \) 2

\(\angle \) BAD = \(75^{\circ} \).

3. In the given figure, ABCD is a cyclic quadrilateral. Calculate the size of angle BCD.

Solution: We have,

• The \(\angle \) BAD = \(51^{\circ} \)
• The \(\angle \) BCD = \(\theta \)

We already knew that the \(\angle \) BAD = \(51^{\circ} \), we do not need to use any other angle fact to determine this angle.

As the opposite angles in a cyclic quadrilateral total \(180^{\circ} \), we can calculate the size of \(\angle \) BCD.

\(\angle \) BAD + \(\angle \) BCD = \(180^{\circ} \)

\(\angle \) BCD = \(180^{\circ} \) – \(\angle \) BAD

\(\angle \) BCD = \(180^{\circ} \) – \(51^{\circ} \)

\(\angle \) BCD = \(129^{\circ} \).

4. In the given figure, points A, B, and C are on the circumference of a circle with centre O. DE is a tangent at point A. Calculate the size of the angle BAD.

Solution: We have,

• The \(\angle \) BCA = \(52^{\circ} \)
• AC is the diameter
• DE is the tangent
• The \(\angle \) BAD = \(\theta \)

As AC is the diameter and the angle in a semicircle is \(90^{\circ} \), \(\angle \) ABC = \(90^{\circ} \). And we know angles in a triangle total \(180^{\circ} \).

\(\angle \) CAB = \(180^{\circ} \) – (\(90^{\circ} \) + \(52^{\circ} \))

\(\angle \) CAB = \(38^{\circ} \)

As the angle between the tangent and the radius is \(90^{\circ} \), we can now calculator \(\angle \) BAD.

\(\angle \) BAD = \(90^{\circ} \) – \(38^{\circ} \)

\(\angle \) BAD = \(52^{\circ} \).

We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.

Q.1. How to prove circle theorems?

Ans.  We can prove circle theorems by using various results in geometry like the triangle sum theorem, theorems based on angles in geometry, etc.

Q.2.  How many theorems are in a circle ?

Ans. There are mainly seven circle theorems:

• Alternate segment circle theorem
• The angle at the centre circle theorem
• Angles in the same segment circle theorem
• Angle in a semi-circle theorem
• Chord circle theorem
• Tangent circle theorem
• Cyclic quadrilateral theorem

Q.3. Why do we need circle theorems?

Ans. We need circle theorems to solve different types of problems in geometry. When we draw angles and lines inside the circle, we can deduce different patterns and results from it which are helpful in practical and real-life scenarios.

Q.4. How are circle theorems used in real life?

Ans. Circle theorems are used in real life in various situations. Some of them are as follows: A wheelmaker needs to know the properties and theorems of a circle. In order for a wheel to function properly, the diameter has to remain exactly the same length. To do this, the wheel maker must have knowledge of the radius as well. Construction workers use theorems of circles when they create stadiums or domes that are circular.

Q.5. How to find angles in circle theorems?

Ans. Angles in circle theorems can be found using the circle theorems based on angles. Some of the circle theorems based on angles are:

• The angle at the centre circle theorem
• Angles in the same segment circle theorem
• Angle in a semi-circle theorem