AREAS RELATED TO CIRCLES-Notes

The chapter “Areas Related to Circles” marks a significant transition from linear measurement to the study of curved surfaces, enriching students’ understanding of geometry beyond straight lines and polygons. Building upon prior knowledge of circles, this chapter systematically introduces methods to calculate the area enclosed by circular boundaries and their parts. Students learn how angles at the centre govern the division of a circle and how proportional reasoning connects arc length, sector area, and the complete circle. This chapter carefully develops the concepts of sector, segment, major and minor regions, and applies them to meaningful real-life contexts such as fields, tracks, wheels, gardens, and decorative designs. It also reinforces algebraic manipulation, unit consistency, and logical problem-solving. By blending geometric intuition with formula-based reasoning, the chapter strengthens analytical skills that are essential for higher mathematics and competitive examinations. Mastery of this topic enables learners to confidently solve both textbook exercises and application-oriented problems involving circular shapes.

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CIRCLES-Notes

The chapter “Circles” in NCERT Mathematics Class X introduces learners to one of the most elegant and logically rich areas of geometry. Building upon earlier concepts of lines, angles, and triangles, this chapter focuses on understanding the geometric behaviour of circles through precise definitions, carefully reasoned theorems, and visual interpretation. Students explore fundamental ideas such as tangents, secants, chords, points of contact, and the unique relationships that exist between a circle and the lines associated with it. A major strength of this chapter lies in its emphasis on logical proofs, where learners develop mathematical reasoning by justifying why a tangent is perpendicular to the radius or why tangents drawn from an external point are equal in length. These results are not only central to geometry but also enhance problem-solving skills required for higher mathematics. The chapter balances theory with application, helping students confidently tackle construction problems, numerical questions, and proof-based questions commonly asked in board examinations. Mastery of this chapter strengthens spatial reasoning, sharpens diagram interpretation skills, and lays a strong foundation for advanced geometry. With consistent practice and conceptual clarity, students can score high marks while appreciating the symmetry and precision that circles bring to mathematics.

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December 15, 2025 | By Academia Aeternum

AREAS RELATED TO CIRCLES-Notes

Maths - Notes

Sector

A major sector of a circle is the larger region formed when two radii of a circle divide it into two unequal parts. When the angle between the two radii at the centre is greater than $180^\circ$, the sector corresponding to this angle is called the major sector. The remaining smaller part, with an angle less than $180^\circ$, is known as the minor sector.

The concept of a major sector is closely related to that of a minor sector, as both together make up the entire circle. Since the total angle around the centre of a circle is $360^\circ$, the angle of the major sector can be found by subtracting the angle of the minor sector from $360^\circ$. Because the area of a sector is directly proportional to its central angle, the major sector always occupies more than half of the area of the circle.

To find the area of a major sector, we usually calculate the area of the whole circle and then subtract the area of the corresponding minor sector. This approach simplifies calculations and avoids dealing with large central angles directly. The formula reflects the idea that the major sector is simply what remains after removing the smaller sector from the circle.

\[\tiny\begin{aligned} \color{blue} \boxed{\boldsymbol{\textbf{Area of the sector of angle }\theta \\= \frac{\theta}{360}\,\times\pi r^{2}}} \end{aligned}\]

\[\tiny\begin{aligned} \color{blue} \boxed{\boldsymbol{\textbf{Length of an arc of a sector of angle }\theta \\= \frac{\theta}{360}\,\times 2\pi r}} \end{aligned}\]

Segment

A segment of a circle is the region formed when a chord divides the circle into two parts. The chord is a straight line joining any two points on the circumference of the circle, and the curved boundary of the segment is the arc corresponding to that chord. Unlike a sector, a segment does not involve radii directly, but it is closely connected to both circular and triangular geometry.

There are two types of segments: the minor segment and the major segment. The smaller region bounded by the chord and the shorter arc is called the minor segment, while the larger region enclosed by the chord and the longer arc is known as the major segment.

To find the area of a segment, we first consider the sector formed by joining the endpoints of the chord to the centre of the circle. The area of the triangle formed by the two radii and the chord is then subtracted from the area of this sector. This difference gives the area of the minor segment. For the major segment, the area is obtained by subtracting the area of the minor segment from the area of the entire circle.

Example-1

Find the area of the sector of a circle with radius 4 cm and an angle of $\angle 30°$. Also, find the area of the corresponding major sector (Use $\pi$ = 3.14).

Solution:

Radius = 4cm
Angle of minor sector = 30°
Hence, Angle of major sector = 360°-30° = 330°

Area of major sector (A)

$$\begin{aligned} A&=\dfrac{\theta }{360^{\circ }}\cdot \pi r^{2}\\\\ &=\dfrac{330^\circ}{360^\circ}\times 3.14\times 4\times 4\\\\ &\approx 46.1\ cm^{2}\end{aligned}$$

Area of the major sector is $\approx 46.1\ cm^{2}$

Example-2

Find the area of the segment AYB shown in Fig. 11.6, if the radius of the circle is 21 cm and $\angle AOB = 120°.$ (Use $\pi = \frac{22}{7} $)

Solution:

Angle of minor sector = 120°
Radius of the circle = 21cm
Area of the sector $(A_s)$

$$\require{cancel}\begin{aligned}A_s&=\dfrac{\theta }{360}\times \pi r^{2}\\ &=\dfrac{120}{360}\times \dfrac{22}{7}\times 21^2 \\ &=\dfrac{1}{\cancelto{1}3}\times \dfrac{22}{\cancelto{1}7}\times \cancelto{\cancelto{1}3}{21}\times 21\\ &=22\times 21\\ &=462\ cm^{2}\end{aligned}$$

Consider triangle OAB
OA B is an isosceles triangle, angle OAD

$$\small\begin{aligned}\angle OAB+\angle OBA&+\angle AOB=180^{\circ }\\ \angle OAB&=2\angle OBA\\ \therefore 2\angle OAB&=180-\angle AOB\\\\ &=\dfrac{180^{\circ }-120}{2}\\ &=60\\ \Rightarrow \angle OAB&=30^{\circ }\\\\ \sin 30^\circ &=\dfrac{OD}{OA}\\ OD&=OA\cdot \sin 30^{^\circ}\\ &=21\times \dfrac{1}{2}\\ &=\dfrac{21}{2}\\\\ \cos 30^{^\circ}&=\dfrac{AD}{OA}\\ AD&=OA\cdot \cos 30^{\circ }\\ AD&=\dfrac{\sqrt{3}}{2}-21\\ AB&=2AD\\ &=\dfrac{\sqrt{3}}{2}\cdot 21\cdot 2\\ &=21\sqrt{3}\\\\ A_{t}&=\dfrac{1}{2}AB\cdot OD\\\\ &=\dfrac{1}{2}\cdot 21\sqrt{3}\cdot \dfrac{21}{2}\\\\ &=\dfrac{441\sqrt{3}}{2\cdot2}\\\\ &=\dfrac{441}{4}\sqrt{3}\end{aligned}$$

Area of segment $A_{sg}$ = Area of sector-Area of triangle OAB

$$\begin{aligned}A_{sg}&=462-\dfrac{441\sqrt{3}}{4}\\\\ &=\dfrac{1848-441\sqrt{3}}{4}\\\\ &=\dfrac{21}{4}\left( 88-21\sqrt{3}\right) \end{aligned}$$

Frequently Asked Questions

A circle is the locus of all points in a plane that are at a fixed distance, called the radius, from a fixed point known as the centre.

The area of a circle is the region enclosed by its circumference and is calculated using the formula $A = \pi r^2$.

$\pi$ is a constant representing the ratio of the circumference of a circle to its diameter, commonly taken as $\frac{22}{7}$ or 3.14.

A sector is the region bounded by two radii and the arc between them.

A minor sector is the smaller sector formed when the central angle is less than $180^\circ$.

A major sector is the larger sector formed when the central angle is greater than $180^\circ$.

The area of a sector is $\frac{\theta}{360^\circ} \times \pi r^2$, where $\theta$ is the central angle.

The angle at the centre determines what fraction of the circle the sector occupies, directly affecting its area.

A segment is the region bounded by a chord of a circle and the corresponding arc.

A minor segment is the smaller region formed between a chord and the corresponding minor arc.

A major segment is the larger region formed between a chord and the corresponding major arc.

Area of minor segment = Area of corresponding sector - Area of the triangle formed by the radii and chord.

Area of major segment = Area of the circle - Area of the minor segment.

A triangle helps remove the straight-line portion inside the sector, leaving only the curved region of the segment.

A chord is a line segment joining any two points on the circumference of a circle.

An arc is a continuous portion of the circumference of a circle.

Both arc length and sector area are proportional to the central angle of the sector.

Areas are measured in square units such as $\text{cm}^2), (\text{m}^2), or (\text{km}^2$.

Yes, since radius is half of the diameter, it can be substituted accordingly.

The area becomes four times because area is proportional to the square of the radius.

It helps in solving problems related to fields, tracks, wheels, gardens, roads, and circular designs.

It extends mensuration concepts from polygons to curved figures.

Numerical problems on sectors, segments, shaded regions, and word problems based on real-life situations.

Only basic geometric tools like compass and ruler are used for diagrams, not for constructions.

Area subtraction and proportional scaling are the main mathematical transformations used.

Because a segment is obtained by removing a triangular portion from a sector.

By keeping units consistent, using correct values of $\pi$, and identifying the correct region.

No, understanding the relationship between angles and areas is essential.

It involves finding the area of specific parts of a circle shown as shaded in a figure.

By drawing diagrams, identifying known values, and applying appropriate formulas step by step.

Diagrams help visualise sectors, segments, and shaded regions accurately.

It helps relate angles to areas and simplifies calculations.

Yes, real-life based circular layouts are often used in case-study problems.

Basic understanding of circles, triangles, and area formulas.

It builds a foundation for advanced geometry and trigonometry involving circles.

It represents the complete angle around the centre of a circle.

By rounding values properly and following standard calculation steps.

Practising a variety of numerical problems and mastering formula application.

Yes, especially involving composite figures and logical reasoning.

Because it combines geometry, algebra, proportionality, and real-life application.

CIRCLES-Notes

Maths - Notes

Secant

Secant is defined as a straight line that intersects a circle at exactly two distinct points. Unlike a tangent, which touches the circle at only one point, a secant passes through the circle, entering it at one point and leaving it at another. These two points of intersection lie on the circumference of the circle.

Thus, a secant can be visualised as a line that cuts across the circle, forming a chord within the circle between the two points where it meets the boundary.

Important Aspects of a Secant

Number of Points of Intersection:
A secant always intersects a circle at two distinct points. This feature uniquely distinguishes it from:
- a tangent, which has one point of contact, and
- a line that does not meet the circle at all, which has no common point with the circle.
Relationship with Chord:
The segment of the secant that lies inside the circle, between the two points of intersection, is a chord of the circle. Hence, every secant contains a chord, but not every chord is a secant.
Position Relative to the Centre:
A secant may or may not pass through the centre of the circle:
- If a secant passes through the centre, it contains the diameter of the circle.
- If it does not pass through the centre, the chord formed is shorter than the diameter.
Comparison with Tangent:
A secant differs from a tangent in both definition and behavior:
- A secant cuts the circle at two points.
- A tangent touches the circle at exactly one point and is perpendicular to the radius at the point of contact.

Tangent to a Circle

A tangent to a circle is a straight line that touches the circle at exactly one point without cutting through it. This point is known as the point of contact. At this point, the tangent just meets the boundary of the circle and then moves away from it, remaining entirely outside the circle on both sides of the point of contact.

Unlike a secant, which intersects the circle at two points, a tangent has only one common point with the circle.

Important Aspects of a Tangent to a Circle

Single Point of Contact:
A tangent touches the circle at one and only one point. Even though the line extends indefinitely in both directions, it never enters the interior of the circle.
Perpendicularity with the Radius:
The most important property of a tangent is that the radius drawn to the point of contact is perpendicular to the tangent. This means the angle between the radius and the tangent at the point of contact is a right angle. This property is frequently used in proofs and numerical problems.
Position Relative to the Circle:
A tangent always lies outside the circle, except at the point of contact. No part of a tangent lies inside the circular region.
Tangents from an External Point:
From a point outside a circle, two tangents can be drawn to the circle. These tangents:
- touch the circle at different points, and
- are equal in length from the external point to their respective points of contact.
Tangent and Chord Relationship:
The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment of the circle. This result is important in solving angle-based questions involving circles.

Theorem-1

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given:

A circle with centre $O$.
A tangent $YX$ touches the circle at point $P$.

To Prove:

The tangent $YX$ is perpendicular to the radius $OP$, that is, \[OP\perp YX\]

Construction:

Take any point $Q$ on the tangent $YX$, such that $Q\ne P$ Join $OQ$.

Proof:

Since $YX$ is a tangent to the circle at point $P$, it touches the circle at exactly one point.
The point $Q$ lies on the tangent but outside the circle.
Therefore, the distance $OQ$ is greater than the radius of the circle.

Now, consider the distances:

$OQ$ is the radius of the circle.
$OQ\gt OP$, since $Q$ lies outside the circle.

Thus, among all line segments drawn from the centre $O$ to the line $YX$, the segment $OP$ is the shortest distance.

We know from geometry that:

The shortest distance from a point to a line is the perpendicular distance.

Since $OP$ is the shortest distance from the centre $O$ to the tangent $YX$, it must be perpendicular to the tangent.

Hence, \[OP\perp YX\]

Conclusion

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

Number of Tangents from a Point on a Circle

Case 1:
There is no tangent to a circle passing through a point lying inside the circle.
Case 2:
There is one and only one tangent to a circle passing through a point lying on the circle.
Case 3:
There are exactly two tangents to a circle through a point lying outside the circle.

Theorem-2

The lengths of tangents drawn from an external point to a circle are equal.

Proof:

Consider a circle with centre $O$.
Let $P$ be a point outside the circle.
and two tangents $PQ,\ PR$ on the circle from $P$ (As in Fig. 10.3)

Construction:

Join $OP,\ OQ$ and $OR$. Then $\angle OQP$ and $\angle ORP$ are right angles

In right triangles $\triangle OQP$ and $\triangle ORP$, \[ \begin{aligned} OQ&=OR\quad\scriptsize\text{(radii of circle)}\\\\ OP&=OP\quad\scriptsize\text{(Common Hypotenuse)}\\\\ \angle OQP&=\angle ORP\quad\scriptsize (90^\circ)\\\\ \therefore \triangle OQP&\cong \triangle OQP\quad\scriptsize\text{(RHS)}\\\\ \Rightarrow PR&=PQ\quad\scriptsize\text{(CPCT)} \end{aligned} \] Also \[\angle OPQ = \angle OPR\quad\scriptsize\text{(CPCT)}\]

Example-1

Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Solution:

Given:

O is the centre of concentric circles
AB is a chord touchingsmaller circle at pt P

To Prove:

AP = PB

Construction:

Join $AO,\ BO$ and $PO$

Proof:

In $\triangle OAP$ and $\triangle OBP$

$$\begin{aligned} OA&=OB\quad\scriptsize\text{(Radii)}\\\\ OP&=OP\quad\scriptsize\text{(Common)}\\\\ \angle P&=\angle P\quad\scriptsize (90^\circ)\\\\ \therefore \triangle OAP&\cong \triangle OBP\\\\ \Rightarrow AP&=BP\quad\scriptsize\text{(CPCT)} \end{aligned}$$ Hence Proved.

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2 \angle OPQ.$

Solution:

Given:

circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (As in Fig. 10.9)

Proof:

$\triangle TPQ$ is an Isosceles triangle
Hence $\angle P =\angle Q$
Let $\angle PTQ = \theta$

$$\begin{aligned}\angle PTQ&=\theta \\\\ \scriptsize\angle PTQ+\angle TPQ+\angle TQP&=180^{\circ }\\ \scriptsize\text{(Angle Sum}&\scriptsize\text{ Property)}\\\\ \angle TPQ&=\angle TQP\\ \scriptsize\text{(Equal Angles of}&\scriptsize\text{ Isosceles Triangles)}\\\\ \theta +2\angle TPQ&=180^{\circ}\\\\ 2\angle TPQ&=180^{\circ}-\theta \\\\ \angle TPQ&=\dfrac{180^{\circ }-\theta }{2}\\\\ &=90^{\circ }-\dfrac{\theta }{2}\\\\ \angle TPQ+\angle OPQ&=90^{\circ}\\ \scriptsize\text{(Angle between tangent }&\scriptsize\text{TP and Radius OP at Point P)}\\\\ \angle OPQ&=90^{\circ }-\angle TPQ\\\\ \angle OPQ&=90^{\circ }-\left( 90^{\circ}-\dfrac{\theta }{2}\right) \\\\ \angle OPQ&=\dfrac{\theta }{2}\\\\ \Rightarrow \angle OPQ&=\dfrac{\angle PTQ}{2}\\\\ \Rightarrow \angle PTQ&=2\angle OPQ\end{aligned}$$ Hence Proved

Example-3

PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see Fig. 10.10). Find the length TP.

Solution:

Perpendicular subtended by centre of circle to the chord bisect the Chord, Therfore, PR = 8/2 = 4 cm Applying pythagoras theorem of Δ OPR

$$\begin{aligned}OP^{2}&=PR^{2}+OR^{2}\\ OR^{2}&=OP^{2}-PR^{2}\\ &=5^{2}-4^{2}\\ &=25-16\\ OR&=\sqrt{9}\\ OR&=3cm\end{aligned}$$

In $\triangle TRP$ and $\triangle PRO$

$$\begin{aligned} TPR+&\angle RPO=90^{\circ }\\ \angle TPR+&\angle PTR=90^{\circ } \\ \Rightarrow \angle RPO&=\angle PTR\\ \angle TPR&=\angle PRO\\ \therefore \Delta TRP&\sim \triangle PRO\quad\scriptsize\text{(AA Criterion)} \\\\ \dfrac{TP}{PO}&=\dfrac{RP}{R0}\\ \\ TP&=\dfrac{RP\cdot PO}{RO}\\\\ &=\dfrac{4.5}{3}\\\\ &=\dfrac{20}{3}\end{aligned}$$

TP can be found by pythagoras theorem
Let $TP=x$,
$TR=y$
In $\triangle TPR$

$$x^{2}=y^{2}+4^{2} \tag{1} $$

Also in $\triangle TPR$

$$\begin{align}\left( y+3\right) ^{2}&=5^{2}+x^{2}\\ x^{2}&=\left( y+3\right) ^{2}-5^{2}\tag{2} \end{align}$$

Equating (1) and (2)

$$\begin{aligned} \left( y+3\right) ^{2}-25&=y^{2}+16\\ y^{2}+6y+9-25-16-y^{2}&=0\\ 6y-32&=0\\ 6y&=32\\ y&=\dfrac{32}{6}\\ &=\dfrac{16}{3} \end{aligned}$$

Substituting value of $y$ in equation-1

$$\begin{aligned}x^{2}&=y^{2}+16\\ x^{2}&=\left( \dfrac{16}{3}\right) ^{2}+16\\ &=\dfrac{256}{9}+16\\ &=\dfrac{256+144}{9}\\ x^{2} &=\dfrac{400}{9}\\ x&=\sqrt{\dfrac{400}{9}}\\ &=\dfrac{20}{3} \end{aligned}$$

Frequently Asked Questions

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

The centre is the fixed point from which all points on the circle are equidistant.

The radius is the line segment joining the centre of the circle to any point on its circumference.

The diameter is a chord passing through the centre of the circle and is twice the radius.

A chord is a line segment joining any two points on the circumference of a circle.

The diameter is the longest chord of a circle.

A secant is a line that intersects a circle at two distinct points.

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

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

A tangent has exactly one common point with the circle.

A secant has exactly two common points with the circle.

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

The angle is always a right angle (90°).

Exactly one tangent can be drawn from a point on the circle.

No tangent can be drawn from a point inside the circle.

Two tangents can be drawn from a point outside the circle.

The lengths of tangents drawn from an external point to a circle are equal.

They form congruent right-angled triangles with equal radii and a common hypotenuse.

No, a tangent always lies outside the circle except at the point of contact.

It is the square of the length of the tangent drawn from the point to the circle.

The radius drawn to the point of contact is perpendicular to the tangent, not the chord.

No, a diameter always passes through the interior of the circle.

No, a chord is only the segment inside the circle, while a secant is the entire line.

No, but every secant contains a chord within the circle.

The chord formed becomes the diameter of the circle.

The angle between a tangent and a chord equals the angle in the opposite arc of the circle.

It is used to find angles formed by tangents and chords in a circle.

Right-angled triangles are formed with the radius perpendicular to the tangent.

The perpendicular distance from the centre to the tangent is the shortest.

Because the perpendicular gives the shortest distance from the centre to the tangent.

No, they meet at the external point and touch the circle at different points.

The internal segment lies inside the circle; the external segment lies outside.

Constructing tangents from an external point and at a point on the circle.

Yes, in wheels, circular tracks, mechanical parts, and optical instruments.

They help establish right angles, congruence, and length relationships.

Yes, tangents help analyze symmetry and rotational properties of circles.

The tangent–radius perpendicularity theorem.

Proofs, constructions, angle finding, and length-based problems.

Yes, especially based on equal tangents from an external point.

Because properties of tangents and secants are visual and diagram-dependent.

Yes, one tangent at each point on the circle.

It is the path traced by a point moving at a constant distance from a fixed point.

It connects geometry, constructions, and logical proofs.

No, conceptual clarity and diagram understanding are essential.

Learn theorems, practice diagrams, and write step-wise proofs clearly.

AREAS RELATED TO CIRCLES-Notes

TRIGONOMETRIC FUNCTIONS-Exercise 3.2

TRIGONOMETRIC FUNCTIONS-Exercise 3.1

CIRCLES-Notes

TRIGONOMETRIC FUNCTIONS-Exercise 3.2

TRIGONOMETRIC FUNCTIONS-Exercise 3.1

AREAS RELATED TO CIRCLES-Notes

Maths - Notes

Sector

Segment

Example-1

Solution:

Area of the major sector is \(\approx 46.1\ cm^{2}\)

Example-2

Solution:

Frequently Asked Questions

1 What is meant by a circle in geometry?

2 What is the area of a circle?

3 What does \(\pi\) represent in circle formulas?

4 Define a sector of a circle.

5 What is a minor sector?

6 What is a major sector?

7 How is the area of a sector calculated?

8 Why is the angle at the centre important in sector problems?

9 What is a segment of a circle?

10 What is a minor segment?

11 What is a major segment?

12 How do you find the area of a minor segment?

13 How is the area of a major segment obtained?

14 Why are triangles used while finding the area of a segment?

15 What is a chord of a circle?

16 What is an arc of a circle?

17 What is the relationship between arc length and sector area?

18 What units are used for area related to circles?

19 Can diameter be used instead of radius in formulas?

20 What happens to the area if the radius is doubled?

21 Why is this chapter important for real-life applications?

22 How is this chapter linked to mensuration?

23 What type of problems are commonly asked in exams from this chapter?

24 Are construction tools used in this chapter?

25 What transformations are used in this chapter?

26 Why is subtraction used in segment problems?

27 How can mistakes be avoided in exam calculations?

28 Is memorisation of formulas enough for this chapter?

29 What is a shaded region problem?

30 How should students approach word problems?

31 Why are diagrams important in this chapter?

32 What role does proportional reasoning play here?

33 Can this chapter appear in case-study questions?

34 What prior knowledge is required for this chapter?

35 How does this chapter help in higher classes?

36 What is the significance of \(360^\circ\) in this chapter?

37 How is accuracy maintained in answers?

38 What is the most scoring strategy for this chapter?

39 Are HOTS questions asked from this chapter?

40 Why is this chapter considered conceptually rich?

Recent posts

CIRCLES-Notes

Maths - Notes

Secant

Important Aspects of a Secant

Tangent to a Circle

Important Aspects of a Tangent to a Circle

Theorem-1

Given:

To Prove:

Construction:

Proof:

Conclusion

Number of Tangents from a Point on a Circle

Theorem-2

Proof:

Construction:

Example-1

Solution:

Given:

To Prove:

Construction:

Proof:

Solution: