Circles Notes

10

Circles

Class 10 Mathematics • Complete Board Exam Notes

Easy English • Step-by-Step Solutions • Theorem Focus • Exam Ready

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Chapter Roadmap

1. Circle Basics 2. Tangent to a Circle 3. Radius ⟂ Tangent Theorem 4. Number of Tangents 5. Equal Tangents Theorem 6. Important Theorem Proofs 7. Applications & Problems 8. Board-Exam Questions 9. Ten Detailed Solved Questions 10. Exam Strategy 11. Quick Revision

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Chapter Goal

This chapter is mainly about tangents to a circle. You will learn what a tangent is, how a tangent is related to the radius at the point of contact, how many tangents can be drawn from a point, and why tangents drawn from the same external point are equal.

1. Circle Basics

1.1 What is a Circle?

A circle is the set of all points in a plane that are at the same distance from a fixed point.

Centre: The fixed point is called the centre of the circle.
Radius: The equal distance from the centre to any point on the circle is called the radius.

1.2 Important Terms

• Centre: fixed point from which every point on the circle is equally distant.
• Radius: line segment joining the centre to any point on the circle.
• Diameter: a chord passing through the centre; its length is \(2r\).
• Chord: line segment joining any two points on a circle.
• Secant: a line that intersects a circle at two points.
• Tangent: a line that touches the circle at exactly one point.
• Point of contact: the point where a tangent touches the circle.

\[ \boxed{\text{Diameter}=2\times\text{Radius}} \qquad \boxed{d=2r} \]

Solved Example 1 — Find Diameter

The radius of a circle is \(7\) cm. Find its diameter.

1. Use: \[ d=2r. \]
2. Substitute \(r=7\): \[ d=2(7)=14. \]

Answer: \(14\) cm

Solved Example 2 — Find Radius

The diameter of a circle is \(18\) cm. Find its radius.

1. Use: \[ r=\frac d2. \]
2. Substitute: \[ r=\frac{18}{2}=9. \]

Answer: \(9\) cm

Solved Example 3 — Identify a Tangent

A line meets a circle at exactly one point \(P\). What is the line called?

1. A line that touches a circle at exactly one point is a tangent.
2. The point \(P\) is the point of contact.

Answer: The line is a tangent to the circle at \(P\).

2. Tangent to a Circle

2.1 Definition of Tangent

A tangent is a line that intersects a circle at exactly one point.

\[ \boxed{\text{A tangent touches a circle at exactly one point.}} \]

2.2 Tangent and Secant

Line	Number of Common Points with Circle
Tangent	Exactly one
Secant	Exactly two

Exam memory: Tangent = one point; Secant = two points.

Solved Example 1 — Tangent or Secant?

A line intersects a circle at points \(A\) and \(B\). Is it a tangent or a secant?

1. The line has two common points with the circle.
2. A line intersecting a circle at two points is a secant.

Answer: Secant

Solved Example 2 — Point of Contact

A tangent touches a circle at \(P\). What is \(P\) called?

1. The single point where a tangent touches a circle is called the point of contact.

Answer: \(P\) is the point of contact.

Solved Example 3 — Recognise a Tangent

A line \(l\) touches a circle with centre \(O\) at \(T\). What geometric fact is immediately associated with \(OT\) and \(l\)?

1. \(OT\) is a radius because \(T\) lies on the circle.
2. The radius through the point of contact is perpendicular to the tangent.

Answer: \(OT\perp l\).

3. Radius Perpendicular to Tangent Theorem

3.1 Main Theorem

\[ \boxed{\text{The tangent at any point of a circle is perpendicular to the radius through the point of contact.}} \] \[ \boxed{OT\perp PT} \]

If \(PT\) is tangent to a circle with centre \(O\) at \(T\), then the angle between \(OT\) and \(PT\) is \(90^\circ\).

\[ \boxed{\angle OTP=90^\circ} \]

3.2 Why This Theorem is Important

This theorem converts a circle problem into a right-triangle problem. Once a \(90^\circ\) angle is known, Pythagoras theorem and trigonometric ratios can often be used.

Solved Example 1 — Find the Angle

A tangent \(PT\) touches a circle at \(T\), and \(OT\) is the radius. Find \(\angle OTP\).

1. Radius through the point of contact is perpendicular to the tangent.
2. Therefore: \[ OT\perp PT. \]
3. Hence: \[ \angle OTP=90^\circ. \]

Answer: \(90^\circ\)

Solved Example 2 — Pythagoras with a Tangent

A circle has radius \(6\) cm. From an external point \(P\), the tangent length \(PT\) is \(8\) cm. Find \(OP\).

1. Since \(OT\perp PT\), triangle \(OTP\) is right-angled at \(T\).
2. Use Pythagoras: \[ OP^2=OT^2+PT^2. \]
3. Substitute: \[ OP^2=6^2+8^2=36+64=100. \]
4. Therefore: \[ OP=10. \]

Answer: \(10\) cm

Solved Example 3 — Find Tangent Length

From a point \(P\), \(OP=13\) cm and the radius \(OT=5\) cm. Find the tangent length \(PT\).

1. Since \(OT\perp PT\), triangle \(OTP\) is right-angled.
2. Use: \[ PT^2=OP^2-OT^2. \]
3. Substitute: \[ PT^2=13^2-5^2=169-25=144. \]
4. Thus: \[ PT=12. \]

Answer: \(12\) cm

4. Number of Tangents

4.1 Position of a Point and Number of Tangents

Position of Point	Number of Tangents
Inside the circle	0
On the circle	1
Outside the circle	2

Important board fact: From an external point, exactly two tangents can be drawn to a circle.

Solved Example 1 — Point Inside

How many tangents can be drawn from a point inside a circle?

1. A point inside the circle cannot have a line touching the circle at exactly one point while remaining in the required tangent position.

Answer: \(0\)

Solved Example 2 — Point on the Circle

How many tangents can be drawn at a point \(T\) lying on the circle?

1. At a point on the circle, exactly one tangent can be drawn.

Answer: \(1\)

Solved Example 3 — External Point

A point \(P\) lies outside a circle. How many tangents can be drawn from \(P\)?

1. For an external point, there are two tangent positions.
2. Let them touch the circle at \(A\) and \(B\).

Answer: \(2\) tangents

5. Equal Tangents Theorem

5.1 Main Theorem

\[ \boxed{\text{The lengths of tangents drawn from an external point to a circle are equal.}} \] \[ \boxed{PA=PB} \]

If \(PA\) and \(PB\) are tangents from an external point \(P\), touching the circle at \(A\) and \(B\), then the two tangent lengths are equal.

5.2 Proof Idea

Join the centre \(O\) to \(A\) and \(B\). Since radii are perpendicular to tangents, triangles \(OAP\) and \(OBP\) are right triangles. They have a common hypotenuse \(OP\) and equal radii \(OA=OB\). Therefore they are congruent by RHS, giving \(PA=PB\).

Solved Example 1 — Direct Equality

From an external point \(P\), tangents \(PA\) and \(PB\) are drawn. If \(PA=9\) cm, find \(PB\).

1. Tangents from the same external point are equal.
2. Therefore: \[ PB=PA=9. \]

Answer: \(9\) cm

Solved Example 2 — Find Unknown

If \(PA=2x+3\) and \(PB=5x-12\) are tangents from \(P\), find \(x\).

1. Equal tangents: \[ PA=PB. \]
2. Therefore: \[ 2x+3=5x-12. \]
3. Rearrange: \[ 15=3x. \]
4. Hence: \[ x=5. \]
5. Tangent length: \[ PA=2(5)+3=13. \]

Answer: \(x=5\), and each tangent is \(13\) units.

Solved Example 3 — Perimeter Application

A circle touches all four sides of a quadrilateral \(ABCD\). If the tangent segments from each vertex are equal, show that \(AB+CD=BC+DA\).

1. Let the tangent lengths from \(A,B,C,D\) be \(x,y,z,w\), respectively.
2. Then: \[ AB=x+y,\quad BC=y+z, \] \[ CD=z+w,\quad DA=w+x. \]
3. Add \(AB+CD\): \[ AB+CD=x+y+z+w. \]
4. Add \(BC+DA\): \[ BC+DA=y+z+w+x. \]
5. Both sums are equal.

Hence, \(AB+CD=BC+DA\).

6. Important Theorem Proofs

6.1 Proof: Radius is Perpendicular to Tangent

Theorem

If \(PT\) is tangent to a circle with centre \(O\) at \(T\), then \(OT\perp PT\).

Proof idea: Suppose the radius \(OT\) were not perpendicular to the tangent. From the centre \(O\), draw a perpendicular \(OM\) to the tangent. Since the perpendicular is the shortest distance from a point to a line, \(OM<OT\). But \(OT\) is the radius, so a point \(M\) on the tangent would lie inside the circle. The line would then meet the circle at more than one point, contradicting that \(PT\) is a tangent. Therefore \(OT\perp PT\).

Hence proved.

6.2 Proof: Tangents from an External Point are Equal

Theorem

From an external point \(P\), tangents \(PA\) and \(PB\) are drawn to a circle with centre \(O\). Prove \(PA=PB\).

Proof:

1. Join \(OA,OB\) and \(OP\).
2. Since radius is perpendicular to tangent: \[ OA\perp PA,\qquad OB\perp PB. \]
3. Thus \(\triangle OAP\) and \(\triangle OBP\) are right triangles.
4. \(OA=OB\), because both are radii of the same circle.
5. \(OP=OP\), common hypotenuse.
6. Therefore: \[ \triangle OAP\cong\triangle OBP \] by RHS congruence.
7. Corresponding sides are equal: \[ PA=PB. \]

Hence proved.

6.3 Converse-Style Reasoning

In numerical questions, if two line segments from the same external point are stated to be tangents to the same circle, immediately write their lengths equal. This often reduces the problem to a simple linear equation.

Solved Example 1 — Linear Equation

Two tangent lengths from \(P\) are \(3x-2\) and \(x+8\). Find \(x\).

1. Equal tangents: \[ 3x-2=x+8. \]
2. Therefore: \[ 2x=10. \]
3. Hence: \[ x=5. \]

Answer: \(x=5\)

Solved Example 2 — Tangent Length

If two tangents from \(P\) are \(4x+1\) and \(7x-14\), find the common tangent length.

1. Set them equal: \[ 4x+1=7x-14. \]
2. So: \[ 15=3x\Rightarrow x=5. \]
3. Common length: \[ 4(5)+1=21. \]

Answer: \(21\) units

Solved Example 3 — Right Triangle Application

From an external point \(P\), \(OP=17\) cm and the radius is \(8\) cm. Find the tangent length.

1. The radius is perpendicular to the tangent.
2. Use Pythagoras: \[ PT^2=OP^2-OT^2. \]
3. Substitute: \[ PT^2=17^2-8^2=289-64=225. \]
4. Thus: \[ PT=15. \]

Answer: \(15\) cm

7. Applications & Problems

7.1 Tangent Length Formula

If \(P\) is outside a circle with centre \(O\), radius \(r\), and \(PT\) is a tangent, then triangle \(OTP\) is right-angled.

\[ \boxed{PT=\sqrt{OP^2-r^2}} \]

Solved Example 1 — Use the Tangent Formula

A point is \(25\) cm from the centre of a circle of radius \(7\) cm. Find the tangent length.

1. Use: \[ PT=\sqrt{OP^2-r^2}. \]
2. Substitute: \[ PT=\sqrt{25^2-7^2}. \]
3. Calculate: \[ PT=\sqrt{625-49}=\sqrt{576}=24. \]

Answer: \(24\) cm

Solved Example 2 — Find Distance from Centre

The radius of a circle is \(9\) cm and a tangent from \(P\) is \(12\) cm. Find \(OP\).

1. Use: \[ OP^2=OT^2+PT^2. \]
2. Substitute: \[ OP^2=9^2+12^2=81+144=225. \]
3. Therefore: \[ OP=15. \]

Answer: \(15\) cm

Solved Example 3 — Check Whether a Tangent is Possible

A circle has radius \(10\) cm. Can a tangent be drawn from a point \(P\) such that \(OP=8\) cm?

1. For a point to have a tangent to the circle, it must lie outside the circle.
2. Here: \[ OP=8<10=r. \]
3. Therefore \(P\) lies inside the circle.

Answer: No tangent can be drawn from \(P\).

8. Board-Exam Questions

Board Pattern 1

Question: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Answer approach: State the theorem and give the contradiction proof using the fact that the perpendicular from a point to a line is the shortest distance.

Key result: \(OT\perp PT\).

Board Pattern 2

Question: From an external point \(P\), two tangents \(PA\) and \(PB\) are drawn to a circle. Prove that \(PA=PB\).

Solution: Join \(OA,OB,OP\). Both triangles are right triangles, \(OA=OB\) and \(OP\) is common. Hence the triangles are congruent by RHS, giving \(PA=PB\).

Hence proved.

Board Pattern 3

Question: A circle has radius \(5\) cm and \(OP=13\) cm. Find the tangent length from \(P\).

\[ PT=\sqrt{OP^2-r^2} =\sqrt{13^2-5^2} =\sqrt{169-25} =12. \]

Answer: \(12\) cm

Board Pattern 4

Question: If \(PA\) and \(PB\) are tangents from \(P\), \(PA=3x+2\) and \(PB=5x-10\). Find \(x\) and the tangent length.

\[ 3x+2=5x-10 \] \[ 12=2x\Rightarrow x=6. \] \[ PA=3(6)+2=20. \]

Answer: \(x=6\), tangent length \(=20\) units.

9. Ten Detailed Solved Questions

Question 1

A circle has radius \(7\) cm. Find its diameter.

1. Use: \[ d=2r. \]
2. Substitute: \[ d=2(7)=14. \]

Answer: \(14\) cm

Question 2

A tangent \(PT\) touches a circle with centre \(O\) at \(T\). If \(OT=9\) cm and \(OP=15\) cm, find \(PT\).

1. Since radius is perpendicular to tangent: \[ OT\perp PT. \]
2. Use Pythagoras: \[ PT^2=OP^2-OT^2. \]
3. Substitute: \[ PT^2=15^2-9^2=225-81=144. \]
4. Therefore: \[ PT=12. \]

Answer: \(12\) cm

Question 3

Two tangents from an external point are \(4x+3\) and \(7x-12\). Find \(x\).

1. Tangents from the same external point are equal: \[ 4x+3=7x-12. \]
2. Rearrange: \[ 15=3x. \]
3. Hence: \[ x=5. \]

Answer: \(x=5\)

Question 4

A point \(P\) is \(17\) cm from the centre of a circle of radius \(8\) cm. Find the length of the tangent from \(P\).

1. Use: \[ PT=\sqrt{OP^2-r^2}. \]
2. Substitute: \[ PT=\sqrt{17^2-8^2} =\sqrt{289-64} =\sqrt{225}. \]
3. Thus: \[ PT=15. \]

Answer: \(15\) cm

Question 5

How many tangents can be drawn from a point at the centre of a circle?

1. The centre lies inside the circle.
2. From a point inside a circle, no tangent can be drawn.

Answer: \(0\)

Question 6

From an external point \(P\), \(PA\) and \(PB\) are tangents. If \(PA=18\) cm, find \(PB\).

1. Use equal tangents theorem: \[ PA=PB. \]
2. Therefore: \[ PB=18. \]

Answer: \(18\) cm

Question 7

Prove that the angle between the radius and tangent at the point of contact is \(90^\circ\).

1. Let \(PT\) be tangent at \(T\) and \(OT\) be the radius.
2. By the radius-tangent theorem: \[ OT\perp PT. \]
3. Perpendicular lines form a right angle: \[ \angle OTP=90^\circ. \]

Hence proved.

Question 8

A circle has radius \(12\) cm. A point \(P\) is \(20\) cm from its centre. Find the tangent length.

1. Use: \[ PT=\sqrt{OP^2-r^2}. \]
2. Substitute: \[ PT=\sqrt{20^2-12^2} =\sqrt{400-144} =\sqrt{256}. \]
3. Therefore: \[ PT=16. \]

Answer: \(16\) cm

Question 9

If two tangents from \(P\) are \(2x+5\) and \(5x-10\), find \(x\) and each tangent length.

1. Equal tangents: \[ 2x+5=5x-10. \]
2. Rearrange: \[ 15=3x\Rightarrow x=5. \]
3. Substitute: \[ PA=2(5)+5=15. \]
4. Check: \[ PB=5(5)-10=15. \]

Answer: \(x=5\), each tangent \(=15\) units.

Question 10

A circle touches the four sides of a quadrilateral \(ABCD\). If \(AB=12\) cm, \(BC=9\) cm and \(CD=7\) cm, find \(AD\).

1. For a quadrilateral circumscribed about a circle: \[ AB+CD=BC+AD. \]
2. Substitute: \[ 12+7=9+AD. \]
3. Therefore: \[ 19=9+AD. \]
4. Hence: \[ AD=10. \]

Answer: \(10\) cm

10. Exam Strategy

1. Memorise the two main theorems.

Radius at the point of contact is perpendicular to the tangent, and tangents from the same external point are equal.

2. Look for a right triangle.

Whenever a tangent and radius meet at the point of contact, immediately mark \(90^\circ\). This can unlock Pythagoras theorem.

3. For algebra questions, write equal tangents first.

If \(PA\) and \(PB\) are tangents from the same external point: \[ PA=PB. \] This often gives a simple linear equation.

4. Do not confuse tangent and secant.

Tangent has exactly one common point with the circle; secant has two common points.

Most Common Errors

• Forgetting that the radius is perpendicular to the tangent at the point of contact.
• Using equal tangents for lines that do not come from the same external point.
• Calling a chord a tangent.
• Forgetting the right angle before applying Pythagoras.
• Using \(OP^2=r^2+PT^2\) in the wrong direction.
• Giving the negative square root when a length is required.
• Skipping theorem statements in proof-based questions.

11. Quick Revision Sheet

Concept	Formula / Key Fact
Radius	Distance from centre to any point on the circle
Diameter	\(d=2r\)
Tangent	Touches circle at exactly one point
Secant	Intersects circle at two points
Radius–Tangent	\(OT\perp PT\)
Right Angle	\(\angle OTP=90^\circ\)
Equal Tangents	\(PA=PB\)
Tangent Length	\(PT=\sqrt{OP^2-r^2}\)
Point Inside	0 tangents
Point on Circle	1 tangent
Point Outside	2 tangents
Equal Tangent Proof	RHS congruence
Circumscribed Quadrilateral	\(AB+CD=BC+AD\)

30-Second Final Revision

• Tangent: one common point.
• Radius + tangent: \(90^\circ\) at point of contact.
• External point: two tangents.
• Equal tangents: \(PA=PB\).
• Tangent length: \(PT=\sqrt{OP^2-r^2}\).
• Proof: use RHS congruence for equal tangents.
• Board answer: theorem → diagram facts → calculation/proof → final statement.

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Class 10 Mathematics • Chapter 10: Circles

Prepared in easy, exam-focused handwritten style by University Scope.

Practise theorem statements, right-triangle applications and equal-tangent questions for maximum accuracy.