Circles | Grade 10th - Mathematics Chapter Summary & NCERT CBSE Study Resources

Study Materials

10th

10th - Mathematics

Circles

Overview of the Chapter: Circles

This chapter introduces the fundamental concepts related to circles, including their properties, theorems, and practical applications. Students will learn about tangents, chords, and various angle properties associated with circles, as prescribed in the CBSE Grade 10 Mathematics curriculum.

Circle: A circle is the collection of all points in a plane that are at a fixed distance from a fixed point in the plane. The fixed point is called the center, and the fixed distance is called the radius.

Key Topics Covered

Tangent to a Circle
Number of Tangents from a Point to a Circle
Lengths of Tangents
Theorems Related to Chords and Angles

Tangent to a Circle

A tangent to a circle is a line that touches the circle at exactly one point. This point is called the point of contact or point of tangency.

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

Number of Tangents from a Point to a Circle

Depending on the position of a point relative to a circle, the number of tangents that can be drawn from the point varies:

If the point lies inside the circle, no tangent can be drawn.
If the point lies on the circle, exactly one tangent can be drawn.
If the point lies outside the circle, exactly two tangents can be drawn.

Lengths of Tangents

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

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

Theorems Related to Chords and Angles

Several important theorems related to chords and angles in circles are covered in this chapter:

The perpendicular from the center of a circle to a chord bisects the chord.
Equal chords of a circle are equidistant from the center.
The angle subtended by an arc at the center is double the angle subtended at any point on the remaining part of the circle.

Theorem: The angle subtended by an arc at the center is double the angle subtended at any point on the remaining part of the circle.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

What is the length of the tangent from a point 5 cm away from the center of a circle with radius 3 cm?

Answer:

4 cm

Question 2:

If two tangents are drawn from an external point to a circle, what can we say about their lengths?

Answer:

They are equal.

Question 3:

What is the angle between two tangents drawn from an external point to a circle of radius 5 cm if the length of each tangent is 12 cm?

Answer:

90°

Question 4:

A point P is 10 cm away from the center of a circle of radius 6 cm. How many tangents can be drawn from P to the circle?

Answer:

Question 5:

What is the perimeter of a quadrant of a circle with radius 14 cm?

Answer:

50 cm

Question 6:

If the angle between two radii of a circle is 120°, what is the angle between the tangents at these points?

Answer:

60°

Question 7:

What is the distance between two parallel tangents of a circle with radius 4 cm?

Answer:

8 cm

Question 8:

If the length of the tangent from a point to a circle is 8 cm and the radius is 6 cm, how far is the point from the center?

Answer:

10 cm

Question 9:

What is the area of a sector of a circle with radius 7 cm and central angle 60°?

Answer:

25.67 cm²

Question 10:

How many common tangents can two circles of radii 5 cm and 3 cm have if the distance between their centers is 10 cm?

Answer:

Question 11:

What is the length of the arc of a sector with radius 10 cm and central angle 45°?

Answer:

7.85 cm

Question 12:

If a quadrilateral ABCD is circumscribed around a circle, what is the relation between AB + CD and AD + BC?

Answer:

AB + CD = AD + BC

Question 13:

Define a circle as per Euclidean geometry.

Answer:

A circle is the set of all points in a plane that are at a fixed distance (called the radius) from a given point (called the center).

Question 14:

What is the circumference of a circle with radius 7 cm? (Use π = 22/7)

Answer:

Circumference = 2πr
Given r = 7 cm
= 2 × (22/7) × 7
= 44 cm

Question 15:

If two circles of radii 5 cm and 3 cm touch externally, what is the distance between their centers?

Answer:

Distance between centers = Sum of radii
= 5 cm + 3 cm
= 8 cm

Question 16:

State the tangent-secant theorem for circles.

Answer:

If a tangent (PT) and a secant (PA) are drawn from an external point P to a circle, then:
PT² = PA × PB
where PB is the external part of the secant.

Question 17:

Calculate the length of a tangent drawn from a point 10 cm away from the center of a circle with radius 6 cm.

Answer:

Using Pythagoras' theorem:
Length of tangent = √(distance² - radius²)
= √(10² - 6²)
= √(100 - 36)
= √64
= 8 cm

Question 18:

What is the angle between two radii of a circle drawn to the endpoints of a diameter?

Answer:

The angle is 180° because the diameter is a straight line and the angle on a straight line is always 180°.

Question 19:

If the angle between two tangents drawn from an external point to a circle is 60°, what is the angle between the radii drawn to the points of contact?

Answer:

The angle between the radii = 180° - angle between tangents
= 180° - 60°
= 120°

Question 20:

How many tangents can be drawn to a circle from a point outside it?

Answer:

Exactly two tangents can be drawn to a circle from an external point.

Question 21:

What is the measure of the angle in a semicircle?

Answer:

The angle in a semicircle is always a right angle (90°). This is known as Thales' theorem.

Question 22:

If two chords of a circle are equal in length, what can you say about their distances from the center?

Answer:

Equal chords are equidistant from the center of the circle. The farther the chord from the center, the smaller its length.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

Define a tangent to a circle and state its key property.

Answer:

A tangent is a line that touches a circle at exactly one point, called the point of tangency.
The key property is that the tangent is perpendicular to the radius at the point of contact.

Question 2:

If the length of a tangent from an external point to a circle is 8 cm and the radius is 6 cm, find the distance of the point from the center.

Answer:

Using the property of tangents:
Distance (d) = √(radius² + tangent length²)
d = √(6² + 8²) = √(36 + 64) = √100 = 10 cm

Question 3:

State the number of tangents that can be drawn from an external point to a circle.

Answer:

Exactly two tangents can be drawn from an external point to a circle.

Question 4:

Prove that the lengths of two tangents drawn from an external point to a circle are equal.

Answer:

Consider two tangents PA and PB from point P to a circle with center O.
In triangles OAP and OBP:
OA = OB (radii)
OP is common
∠OAP = ∠OBP = 90° (tangent ⊥ radius)
∴ ΔOAP ≅ ΔOBP (RHS congruency)
Hence, PA = PB.

Question 5:

If two concentric circles have radii 5 cm and 3 cm, find the length of the chord of the larger circle which touches the smaller circle.

Answer:

Let AB be the chord of the larger circle touching the smaller circle at P.
OP ⊥ AB (radius ⊥ tangent)
In right ΔOAP:
OA = 5 cm (radius of larger circle)
OP = 3 cm (radius of smaller circle)
AP = √(OA² - OP²) = √(25 - 9) = 4 cm
∴ AB = 2 × AP = 8 cm

Question 6:

What is the angle between two tangents drawn from an external point to a circle of radius 5 cm, if the length of each tangent is 12 cm?

Answer:

Let the angle between tangents be θ.
Using the formula: tan(θ/2) = radius/tangent length
tan(θ/2) = 5/12 ⇒ θ/2 = tan⁻¹(5/12)
∴ θ = 2 × tan⁻¹(5/12) ≈ 45.24°

Question 7:

A quadrilateral ABCD is drawn to circumscribe a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, find AD.

Answer:

For a circumscribed quadrilateral, sum of one pair of opposite sides equals the other pair.
AB + CD = AD + BC
6 + 4 = AD + 7
AD = 10 - 7 = 3 cm

Question 8:

State the alternate segment theorem for circles.

Answer:

The alternate segment theorem states that:
The angle between a tangent and a chord is equal to the angle in the alternate segment.
i.e., ∠PTQ = ∠PRQ where PT is tangent and QR is chord.

Question 9:

Two circles touch externally. The sum of their radii is 14 cm and the distance between their centers is 10 cm. Find their radii.

Answer:

For externally touching circles:
Distance between centers = sum of radii
Let radii be r₁ and r₂
r₁ + r₂ = 14 cm
But given distance = 10 cm ≠ 14 cm
This is a contradiction - no such circles exist with these conditions.

Question 10:

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

In a circumscribed parallelogram:
1. Opposite sides are equal (parallelogram property)
2. Sum of adjacent sides are equal (circumscribed quadrilateral property)
Thus, all sides must be equal
∴ It is a rhombus.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Answer:

Let P be an external point, and PA and PB be two tangents drawn from P to a circle with center O.
Join OP, OA, and OB.
In triangles OAP and OBP:
1. OA = OB (radii of the same circle)
2. OP is common
3. ∠OAP = ∠OBP = 90° (tangents are perpendicular to the radius at the point of contact)
Thus, by RHS congruence, ΔOAP ≅ ΔOBP.
Hence, PA = PB (by CPCT).

Question 2:

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point on the minor arc.

Answer:

Let AB be a chord equal to the radius r of the circle with center O.
Join OA and OB to form ΔOAB.
Since OA = OB = AB = r, ΔOAB is an equilateral triangle.
Thus, ∠AOB = 60°.
Now, the angle subtended by chord AB at any point P on the minor arc is half the angle subtended at the center.
So, ∠APB = ½ ∠AOB = 30°.

Question 3:

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer:

Let O be the center of both circles.
Let AB be the chord of the larger circle that touches the smaller circle at P.
Since OP is perpendicular to AB (radius ⊥ tangent at point of contact), ΔOPA is a right triangle.
Using the Pythagoras theorem:
OA² = OP² + AP²
5² = 3² + AP²
25 = 9 + AP²
AP² = 16
AP = 4 cm
Since AB = 2 × AP (perpendicular from center bisects the chord),
AB = 8 cm.

Question 4:

In the given figure, PQ is a tangent to a circle with center O at point P. If ∠OPQ = 50°, find ∠PRQ, where R is a point on the major arc PQ.

Answer:

Since PQ is a tangent and OP is the radius, ∠OPQ = 90°.
Given ∠OPQ = 50°, this implies the angle between the tangent and the chord is 50°.
The angle subtended by chord PQ at the center is twice the angle subtended at any point on the major arc.
Thus, ∠POQ = 2 × ∠PRQ.
But ∠POQ = 180° - 2 × 50° = 80° (since ΔOPQ is a triangle).
Therefore, ∠PRQ = ½ × 80° = 40°.

Question 5:

If two circles intersect at two points, prove that their centers lie on the perpendicular bisector of the common chord.

Answer:

Let two circles with centers O and O' intersect at points A and B.
The common chord is AB.
In triangles OAO' and OBO':
OA = OB (radii of the first circle)
O'A = O'B (radii of the second circle)
OO' = OO' (common side)
Thus, by SSS congruence rule, △OAO' ≅ △OBO'.
Therefore, ∠AOO' = ∠BOO' (by CPCT).
This means OO' is the angle bisector of ∠AOB.
Since OA = OB, △OAB is isosceles, and the angle bisector OO' is also the perpendicular bisector of AB.
Hence, the centers O and O' lie on the perpendicular bisector of the common chord AB.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Use a diagram and logical steps.

Answer:

Introduction

We studied that a tangent touches the circle at exactly one point. Our textbook shows that this tangent is always perpendicular to the radius at the point of contact.

Argument 1

Let’s assume the tangent is not perpendicular to the radius.
Draw a perpendicular from the center to the tangent, meeting at point P.

Argument 2

The shortest distance from the center to the tangent is the perpendicular.
But radius is already the shortest distance, so tangent must be perpendicular.

Conclusion

Hence, the tangent is perpendicular to the radius at the point of contact. [Diagram: Circle with tangent and radius labeled]

Question 2:

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Show steps.

Answer:

Introduction

Our textbook explains that concentric circles have the same center. Here, we need to find the chord of the larger circle tangent to the smaller one.

Argument 1

Draw the two circles with center O. Let AB be the chord of the larger circle touching the smaller circle at P.
OP is perpendicular to AB (as tangent is perpendicular to radius).

Argument 2

In right triangle OPA, OA = 5 cm (radius), OP = 3 cm (radius).
Using Pythagoras, AP = √(5² - 3²) = 4 cm. So, AB = 2 × AP = 8 cm.

Conclusion

The length of the chord is 8 cm. [Diagram: Two concentric circles with chord AB]

Question 3:

Prove that the length of tangents drawn from an external point to a circle are equal. Use a diagram and follow step notation.

Answer:

Introduction

We studied that a tangent is perpendicular to the radius at the point of contact. Let’s prove the given statement using this property.

Argument 1

1. Draw a circle with center O and an external point P.
2. Draw two tangents PA and PB meeting the circle at A and B.
3. Join OA, OB, and OP.

Argument 2

4. In ∆OAP and ∆OBP:
- OA = OB (radii)
- ∠OAP = ∠OBP = 90° (tangent ⊥ radius)
- OP is common.
5. Thus, ∆OAP ≅ ∆OBP (RHS).
6. Hence, PA = PB (CPCT).

Conclusion

Therefore, tangents from an external point are equal. [Diagram: Circle with tangents PA and PB from point P.]

Question 4:

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point on the minor arc. Show steps as per NCERT examples.

Answer:

Introduction

Our textbook shows that the angle subtended by a chord at the center is twice the angle at any point on the circle. Let’s solve this problem stepwise.

Argument 1

1. Let AB be the chord equal to radius r.
2. Draw OA and OB (both radii).
3. Thus, ∆OAB is equilateral, so ∠AOB = 60°.

Argument 2

4. Let P be a point on the minor arc AB.
5. By the theorem, ∠APB = ½ ∠AOB = ½ × 60° = 30°.

Conclusion

Hence, the angle subtended by the chord at the minor arc is 30°. [Diagram: Circle with chord AB = radius and point P on minor arc.]

Question 5:

Prove that the lengths of tangents drawn from an external point to a circle are equal. Use a diagram and logical steps.

Answer:

Introduction

We studied that a tangent is a line touching a circle at exactly one point. Our textbook shows that two tangents from an external point are equal in length.

Argument 1

Draw a circle with center O and an external point P.
Draw two tangents PA and PB touching the circle at A and B.

Argument 2

Join OA, OB, and OP.
Triangles OAP and OBP are congruent by RHS (OA = OB, OP common, ∠OAP = ∠OBP = 90°).
Thus, PA = PB.

Conclusion

Hence, the lengths of tangents from an external point to a circle are equal, as proved.

Question 6:

A chord of a circle is 8 cm long and its distance from the center is 3 cm. Find the radius of the circle using the Pythagoras theorem.

Answer:

Introduction

Our textbook explains that the perpendicular from the center to a chord bisects it. Here, we apply this property to find the radius.

Argument 1

Let AB = 8 cm be the chord, and OM = 3 cm be the perpendicular distance from center O.
Since OM bisects AB, AM = MB = 4 cm.

Argument 2

In right triangle OMA, by Pythagoras theorem: OA² = OM² + AM².
OA² = 3² + 4² = 9 + 16 = 25.
Thus, OA = 5 cm (radius).

Conclusion

Therefore, the radius of the circle is 5 cm, derived using the Pythagoras theorem.

Question 7:

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Show steps and units.

Answer:

Introduction

We studied concentric circles and chords. Our textbook shows that a chord tangent to a smaller circle can be found using the Pythagorean theorem.

Argument 1

Let O be the center. Draw the larger circle (radius = 5 cm) and smaller circle (radius = 3 cm).
The chord AB of the larger circle touches the smaller circle at P, so OP ⊥ AB.

Argument 2

In ΔOAP, OA = 5 cm (radius), OP = 3 cm.
Using Pythagoras, AP = √(5² - 3²) = 4 cm. Thus, AB = 2 × AP = 8 cm.

Conclusion

The length of the chord is 8 cm. [Diagram: Two concentric circles with chord AB tangent to the smaller circle.]

Question 8:

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point on the minor arc. Show steps.

Answer:

Introduction

Our textbook shows that chords and angles are related in circles. Here, the chord equals the radius, forming an equilateral triangle.

Argument 1

Draw the chord AB and radii OA and OB. Since AB = OA = OB, ∆OAB is equilateral.
Thus, ∠AOB = 60° (each angle in an equilateral triangle).

Argument 2

The angle subtended by AB at any point C on the minor arc is half the angle at the center.
So, ∠ACB = ½ × ∠AOB = 30°.

Conclusion

Therefore, the angle subtended by the chord at the minor arc is 30°. [Diagram: Circle with labeled points and angles]

Question 9:

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Show all steps.

Answer:

Introduction

Our textbook explains that concentric circles share the same center. Here, we solve a problem involving tangents and chords.

Argument 1

Let O be the center. The chord AB of the larger circle touches the smaller circle at P.
OP is perpendicular to AB (from tangent property), forming two right triangles OPA and OPB.

Argument 2

Using Pythagoras’ theorem in ΔOPA: OA² = OP² + AP².
Given OA = 5 cm, OP = 3 cm, so AP = 4 cm. Thus, AB = 2 × AP = 8 cm.

Conclusion

The length of the chord is 8 cm. [Diagram: Two concentric circles with chord AB tangent to the smaller circle.]

Question 10:

A circular park of radius 20m has a path of width 2m around it. Find the area of the path. (Use π = 3.14)

Answer:

Introduction

We studied how to find the area of a circular path around a park. Our textbook shows similar problems where we subtract the inner circle’s area from the outer circle.

Argument 1

Inner radius (r) = 20m, width of path = 2m.
Outer radius (R) = 20m + 2m = 22m.

Argument 2

Area of outer circle = πR² = 3.14 × 22 × 22 = 1519.76 m².
Area of inner circle = πr² = 3.14 × 20 × 20 = 1256 m².

Conclusion

Area of path = 1519.76 - 1256 = 263.76 m². This matches NCERT examples on circular paths.

Question 11:

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Show steps.

Answer:

Introduction

Our textbook explains that a chord touching a smaller concentric circle is bisected by it. Here, we apply this concept.

Argument 1

Let the larger circle have radius 5 cm (OA) and the smaller one 3 cm (OM).
Since OM is perpendicular to chord AB, ∆OMA is a right triangle.

Argument 2

Using Pythagoras’ theorem: AM² = OA² - OM² = 25 - 9 = 16.
Thus, AM = 4 cm. Since OM bisects AB, AB = 2 × AM = 8 cm.

Conclusion

The length of the chord is 8 cm. [Diagram: Two concentric circles with chord AB and perpendicular OM.]

Question 12:

A circular park of radius 20m has a road 5m wide running around it. Calculate the area of the road. (Use π = 3.14)

Answer:

Introduction

Our textbook explains how to find the area of a ring-shaped region, like a road around a park.

Argument 1

Inner radius (park) = 20m, outer radius (park + road) = 20m + 5m = 25m.
Area of outer circle = π × (25)² = 3.14 × 625 = 1962.5 m².

Argument 2

Area of inner circle = π × (20)² = 3.14 × 400 = 1256 m².
Road area = Outer area – Inner area = 1962.5 – 1256 = 706.5 m².

Conclusion

The road’s area is 706.5 m². This method applies to real-life problems like calculating pathways.

Question 13:

Prove that the length of tangents drawn from an external point to a circle are equal. Use this property to find the length of tangent PT if OP = 10 cm and radius = 6 cm.

Answer:

Introduction

We studied that tangents from an external point to a circle are equal in length. This is proven using the Right-Angled Triangle concept.

Argument 1

Let’s consider two tangents, PT and PS, from point P to a circle with center O. In ΔOPT and ΔOPS, OP is common, OT = OS (radii), and ∠OTP = ∠OSP = 90°. Thus, ΔOPT ≅ ΔOPS (RHS), proving PT = PS.

Argument 2

Given OP = 10 cm, radius OT = 6 cm. Using Pythagoras’ theorem in ΔOPT: PT = √(OP² - OT²) = √(100 - 36) = 8 cm.

Conclusion

Hence, the length of tangent PT is 8 cm, and the property is verified.

Question 14:

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. If AB = 6 cm, BC = 7 cm, and CD = 4 cm, find AD.

Answer:

Introduction

Our textbook shows that for a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the other pair. This is derived from tangent properties.

Argument 1

Let the circle touch AB, BC, CD, and AD at P, Q, R, S respectively. From tangent properties, AP = AS, BP = BQ, CQ = CR, and DR = DS. Adding these: AP + BP + CR + DR = AS + BQ + CQ + DS ⇒ AB + CD = AD + BC.

Argument 2

Given AB = 6 cm, BC = 7 cm, CD = 4 cm. Using AB + CD = AD + BC: 6 + 4 = AD + 7 ⇒ AD = 3 cm.

Conclusion

Thus, AD is 3 cm, and the property is validated.

Question 15:

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Show steps and units.

Answer:

Introduction

Our textbook explains that concentric circles share the same center. Here, we find the chord of the larger circle tangent to the smaller one.

Argument 1

Draw the figure: Two circles with center O, radii 5 cm (larger) and 3 cm (smaller).
Let AB be the chord of the larger circle touching the smaller circle at P.

Argument 2

OP is perpendicular to AB (radius ⊥ tangent).
In ΔOPA, OA = 5 cm, OP = 3 cm. Using Pythagoras, AP = √(5² - 3²) = 4 cm.

Conclusion

Thus, AB = 2 × AP = 8 cm. [Diagram: Concentric circles with chord AB and perpendicular OP.] Units: cm.

Question 16:

A circular park of radius 20m has a path of width 2m around it. Calculate the area of the path. (Use π = 3.14)

Answer:

Introduction

We learned to find areas of circular paths, similar to NCERT Example 12.2. Here, we solve for a 2m wide path.

Argument 1

Inner radius (r) = 20m, outer radius (R) = 20m + 2m = 22m.
Area of outer circle = πR² = 3.14 × (22)² = 1519.76 m².

Argument 2

Area of inner circle = πr² = 3.14 × (20)² = 1256 m².
Path area = Outer area – Inner area = 1519.76 – 1256 = 263.76 m².

Conclusion

The path area is 263.76 m², calculated using basic circle area formulas. Units are crucial in such real-life problems.

Question 17:

Prove that the lengths of tangents drawn from an external point to a circle are equal. Support your answer with a diagram and step-by-step reasoning.

Answer:

To prove that the lengths of tangents drawn from an external point to a circle are equal, let us consider the following steps:

Given: A circle with center O and an external point P. Two tangents PA and PB are drawn from P to the circle, touching it at points A and B respectively.
To Prove: PA = PB.

Construction: Join OA, OB, and OP.

Proof:
1. In triangles OAP and OBP:
- OA = OB (Radii of the same circle).
- OP = OP (Common side).
- ∠OAP = ∠OBP = 90° (Tangent is perpendicular to the radius at the point of contact).

2. By the RHS (Right Angle-Hypotenuse-Side) congruence rule, △OAP ≅ △OBP.

3. Therefore, by CPCT (Corresponding Parts of Congruent Triangles), PA = PB.

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

Diagram: (Draw a circle with center O, an external point P, and two tangents PA and PB touching the circle at A and B respectively. Highlight the right angles and congruent triangles.)

Value-added information: This property is useful in solving problems related to circles, such as finding distances or proving other geometric properties. It also applies to real-world scenarios, like determining equal distances from a point to a circular boundary.

Question 18:

Prove that the lengths of tangents drawn from an external point to a circle are equal. Use a diagram and provide a step-by-step proof.

Answer:

To prove that the lengths of tangents drawn from an external point to a circle are equal, let's follow these steps:

Hence, the lengths of the two tangents drawn from an external point to a circle are equal.

Question 19:

Prove that the lengths of tangents drawn from an external point to a circle are equal. Also, explain the practical application of this property in real-life scenarios.

Answer:

To prove that the lengths of tangents drawn from an external point to a circle are equal, let us consider the following steps:

Given: A circle with center O and an external point P. Two tangents PA and PB are drawn from P to the circle, touching it at points A and B respectively.

To Prove: PA = PB

Proof:

1. Join OA, OB, and OP.
2. Since PA is a tangent at A, OA is perpendicular to PA (by the tangent-radius property).
3. Similarly, OB is perpendicular to PB.
4. In right triangles OAP and OBP:
- OA = OB (radii of the same circle)
- OP = OP (common side)
5. By the RHS congruence rule, ΔOAP ≅ ΔOBP.
6. Therefore, PA = PB (by CPCT).

Practical Application: This property is used in constructing equal-length supports in architecture, ensuring symmetrical designs in bridges, and even in designing gears where equal force distribution is required. It also helps in navigation and surveying to maintain equal distances from a central point.

Question 20:

Prove that the lengths of tangents drawn from an external point to a circle are equal. Support your answer with a diagram and step-by-step proof.

Answer:

To prove that the lengths of tangents drawn from an external point to a circle are equal, let us consider the following:

Given: A circle with center O and an external point P. Two tangents PA and PB are drawn from P to the circle, touching it at points A and B respectively.
To Prove: PA = PB.

Construction: Join OA, OB, and OP.

Proof:
1. In triangles OAP and OBP:
- OA = OB (Radii of the same circle).
- OP = OP (Common side).
- ∠OAP = ∠OBP = 90° (Tangent is perpendicular to the radius at the point of contact).

2. By the RHS (Right Angle-Hypotenuse-Side) congruence rule, △OAP ≅ △OBP.

3. Therefore, by CPCT (Corresponding Parts of Congruent Triangles), PA = PB.

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

Diagram: (Draw a circle with center O, an external point P, and two tangents PA and PB touching the circle at A and B. Label all points and indicate right angles at A and B.)

Value-Added Information: This property is useful in solving problems related to circles, such as finding unknown lengths or proving other geometric properties. It also applies to real-world scenarios, like determining equal distances from a point to a circular boundary.

Question 21:

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. Support your answer with a diagram and detailed reasoning.

Answer:

To prove that AB + CD = AD + BC for a quadrilateral ABCD circumscribing a circle, we use the property of tangents from an external point:

Given: A quadrilateral ABCD circumscribes a circle, touching it at points P, Q, R, and S on sides AB, BC, CD, and DA respectively.
To Prove: AB + CD = AD + BC.
Proof:
1. From point A, the lengths of the two tangents are equal:
- AP = AS ...(1)
2. From point B, the lengths of the two tangents are equal:
- BP = BQ ...(2)
3. From point C, the lengths of the two tangents are equal:
- CR = CQ ...(3)
4. From point D, the lengths of the two tangents are equal:
- DR = DS ...(4)
5. Adding equations (1), (2), (3), and (4):
- AP + BP + CR + DR = AS + BQ + CQ + DS
6. Simplifying:
- (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
- AB + CD = AD + BC.

Thus, the sum of one pair of opposite sides of a quadrilateral circumscribing a circle is equal to the sum of the other pair.

Question 22:

Prove that the lengths of tangents drawn from an external point to a circle are equal. Using this theorem, find the length of tangent PT if OP = 10 cm and radius of the circle is 6 cm.

Answer:

To prove that the lengths of tangents drawn from an external point to a circle are equal, we consider the following steps:

Given: A circle with center O and an external point P. PA and PB are two tangents to the circle from P.
To Prove: PA = PB.
Proof:
1. Join OA, OB, and OP.
2. In triangles OAP and OBP:
- OA = OB (radii of the same circle)
- OP = OP (common side)
- ∠OAP = ∠OBP = 90° (tangent is perpendicular to radius at the point of contact)
3. By RHS congruency, ΔOAP ≅ ΔOBP.
4. Hence, PA = PB (by CPCT).

Now, to find the length of tangent PT when OP = 10 cm and radius = 6 cm:

1. Since PT is a tangent, ∠OTP = 90°.
2. Using Pythagoras theorem in ΔOTP:
PT² + OT² = OP²
PT² + 6² = 10²
PT² + 36 = 100
PT² = 64
PT = 8 cm.

Thus, the length of the tangent PT is 8 cm.

Question 23:

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Also, justify your answer with a diagram.

Answer:

To find the length of the chord of the larger circle that touches the smaller circle, we follow these steps:

Given: Two concentric circles with radii 5 cm (larger circle) and 3 cm (smaller circle).
To Find: Length of chord AB of the larger circle that touches the smaller circle at point P.
Solution:
1. Draw the diagram with both circles sharing the same center O.
2. Let AB be the chord of the larger circle that touches the smaller circle at P.
3. Since AB is a tangent to the smaller circle, OP is perpendicular to AB (radius perpendicular to tangent at the point of contact).
4. Thus, ΔOPA is a right-angled triangle with OA = 5 cm (radius of larger circle) and OP = 3 cm (radius of smaller circle).
5. Using Pythagoras theorem in ΔOPA:
AP² + OP² = OA²
AP² + 3² = 5²
AP² + 9 = 25
AP² = 16
AP = 4 cm.
6. Since OP is perpendicular to AB, it bisects AB (perpendicular from the center to a chord bisects the chord).
7. Therefore, AB = 2 × AP = 2 × 4 = 8 cm.

The length of the chord AB is 8 cm.

Diagram Justification:
1. The diagram shows two concentric circles with center O.
2. The chord AB of the larger circle touches the smaller circle at P.
3. OP is perpendicular to AB, confirming the application of the tangent and chord properties.

Question 24:

Prove that the lengths of tangents drawn from an external point to a circle are equal. Using this property, solve: Two tangents PA and PB are drawn from an external point P to a circle with center O. If ∠APB = 60°, find the measure of ∠OAB.

Answer:

To prove that the lengths of tangents drawn from an external point to a circle are equal, consider the following steps:

Proof:
1. Let P be an external point, and PA and PB be two tangents to the circle with center O.
2. Join OA, OB, and OP.
3. In triangles OAP and OBP:
- OA = OB (radii of the same circle)
- OP = OP (common side)
- ∠OAP = ∠OBP = 90° (tangent is perpendicular to radius)
4. By RHS congruency, △OAP ≅ △OBP.
5. Hence, PA = PB (by CPCT).

Solution to the problem:
1. Given: PA and PB are tangents, and ∠APB = 60°.
2. Since PA = PB, △PAB is an isosceles triangle.
3. Thus, ∠PAB = ∠PBA = (180° - 60°)/2 = 60°.
4. Since OA is perpendicular to PA, ∠OAP = 90°.
5. Therefore, ∠OAB = ∠OAP - ∠PAB = 90° - 60° = 30°.

Thus, the measure of ∠OAB is 30°.

Question 25:

A circle touches all four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA. Also, if AB = 6 cm, BC = 7 cm, and CD = 4 cm, find the length of DA.

Answer:

To prove that AB + CD = BC + DA for a quadrilateral ABCD circumscribed around a circle, follow these steps:

Proof:
1. Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.
2. By the property of tangents from an external point:
- AP = AS
- BP = BQ
- CQ = CR
- DR = DS
3. Adding these equations:
AP + BP + CR + DR = AS + BQ + CQ + DS
4. Simplifying:
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC

Solution to the problem:
1. Given: AB = 6 cm, BC = 7 cm, CD = 4 cm.
2. From the above proof, AB + CD = BC + DA.
3. Substituting the values:
6 + 4 = 7 + DA
10 = 7 + DA
4. Thus, DA = 10 - 7 = 3 cm.

Therefore, the length of DA is 3 cm.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A circular park of radius 20 m has a path of width 5 m around it. Find the area of the path. (Use π = 3.14)

Answer:

Problem Interpretation

We need to find the area of the path surrounding the circular park.

Mathematical Modeling

Outer radius (R) = 20 m + 5 m = 25 m. Inner radius (r) = 20 m.

Solution

Area of path = π(R² - r²) = 3.14 × (625 - 400) = 3.14 × 225 = 706.5 m².

Question 2:

Two concentric circles have radii 7 cm and 14 cm. Calculate the area of the ring-shaped region between them.

Answer:

Problem Interpretation

We studied concentric circles in our textbook. Here, we find the area between two circles.

Mathematical Modeling

Outer radius (R) = 14 cm, Inner radius (r) = 7 cm.

Solution

Area of ring = π(R² - r²) = (22/7) × (196 - 49) = (22/7) × 147 = 462 cm².

Question 3:

A circular park of radius 20 m has a path of width 5 m around it. Find the area of the path. (Use π = 3.14)

Answer:

Problem Interpretation

We need to find the area of the path surrounding the circular park. The park has a radius of 20 m, and the path is 5 m wide.

Mathematical Modeling

We studied that the area of a ring (path) is the difference between the areas of two concentric circles. Here, the outer radius (R) = 20 m + 5 m = 25 m, and the inner radius (r) = 20 m.

Solution

Area of path = π(R² - r²) = 3.14 × (25² - 20²) = 3.14 × (625 - 400) = 706.5 m².

Question 4:

Two tangents PA and PB are drawn to a circle with center O from an external point P. Prove that PA = PB.

Answer:

Problem Interpretation

We need to prove that the lengths of two tangents (PA and PB) drawn from an external point P to a circle are equal.

Mathematical Modeling

Our textbook shows that tangents from an external point to a circle are equal. We can use the properties of right triangles formed by the radius and tangent.

Solution

Join OA, OB, and OP. In ΔOAP and ΔOBP:

OA = OB (radii)
∠OAP = ∠OBP = 90° (tangent ⊥ radius)
OP is common.

Thus, ΔOAP ≅ ΔOBP (RHS), so PA = PB.

Question 5:

A circular park of radius 20 m has a road 5 m wide running around it. Find the area of the road. (Use π = 3.14)

Answer:

Problem Interpretation

We need to find the area of the road surrounding the circular park. The park has a radius of 20 m, and the road is 5 m wide.

Mathematical Modeling

The road forms a concentric circle around the park. Outer radius (R) = 20 m + 5 m = 25 m.

Solution

Area of park = πr² = 3.14 × (20)² = 1256 m²
Area of park + road = πR² = 3.14 × (25)² = 1962.5 m²
Road area = 1962.5 - 1256 = 706.5 m²

Question 6:

Two concentric circles have radii 7 cm and 14 cm. Find the area of the shaded region between them. (Use π = 22/7)

Answer:

Problem Interpretation

We studied concentric circles in our textbook. Here, we have two circles with radii 7 cm and 14 cm, and we need the area between them.

Mathematical Modeling

The shaded region is the difference in areas of the larger and smaller circles.

Solution

Area of smaller circle = πr² = (22/7) × 7 × 7 = 154 cm²
Area of larger circle = πR² = (22/7) × 14 × 14 = 616 cm²
Shaded area = 616 - 154 = 462 cm²

Question 7:

A circular park of radius 20 m has a road 5 m wide running around it. Calculate the area of the road. (Use π = 3.14)

Answer:

Problem Interpretation

We need to find the area of the road surrounding the circular park. The park has radius 20 m, and the road is 5 m wide.

Mathematical Modeling

Outer radius (R) = Park radius + Road width = 20 m + 5 m = 25 m.

Solution

Area of road = πR² - πr² = 3.14 × (25² - 20²) = 3.14 × (625 - 400) = 706.5 m².

Question 8:

Two concentric circles have radii 7 cm and 14 cm. Find the area of the shaded region between them.

Answer:

Problem Interpretation

We studied concentric circles in our textbook. Here, we have two circles with radii 7 cm and 14 cm sharing the same center.

Mathematical Modeling

Shaded area = Area of larger circle - Area of smaller circle.

Solution

Area = π(14² - 7²) = (22/7) × (196 - 49) = (22/7) × 147 = 462 cm².

Question 9:

Two concentric circles have radii 7 cm and 14 cm. Calculate the area of the shaded region between them. (Use π = 22/7)

Answer:

Problem Interpretation

We studied concentric circles in our textbook. Here, two circles share the same center with radii 7 cm and 14 cm.

Mathematical Modeling

Shaded area = Area of larger circle - Area of smaller circle.

Solution

Area = π(14² - 7²) = (22/7) × (196 - 49) = (22/7) × 147 = 462 cm².

Question 10:

Two concentric circles have radii 7 cm and 14 cm. Calculate the area of the shaded region between them. (Use π = 22/7)

Answer:

Problem Interpretation

We studied concentric circles in our textbook. Here, we have two circles with radii 7 cm and 14 cm sharing the same center.

Mathematical Modeling

The shaded region is the area between the two circles. We subtract the area of the smaller circle from the larger one.

Solution

Area of larger circle = πR² = (22/7) × (14)² = 616 cm²
Area of smaller circle = πr² = (22/7) × (7)² = 154 cm²
Shaded area = 616 - 154 = 462 cm²

Question 11:

In a park, a circular fountain is surrounded by a pathway of uniform width 2 meters. The radius of the fountain is 7 meters. A gardener wants to plant flowers along the outer edge of the pathway. Calculate the length of the outer edge where the flowers will be planted. Also, find the area of the pathway.

Answer:

Given: Radius of fountain (r) = 7 m, Width of pathway (w) = 2 m.

Step 1: Calculate the radius of the outer edge (including pathway).
Outer radius (R) = r + w = 7 m + 2 m = 9 m.

Step 2: Find the circumference of the outer edge (where flowers are planted).
Circumference = 2πR = 2 × 3.14 × 9 ≈ 56.52 m.

Step 3: Calculate the area of the pathway.
Area of pathway = π(R² - r²) = 3.14 × (81 - 49) ≈ 100.48 m².

Final Answer: The length of the outer edge is 56.52 meters, and the area of the pathway is 100.48 square meters.

Question 12:

Two concentric circles have radii 5 cm and 13 cm. A chord of the larger circle touches the smaller circle. Find the length of this chord.

Answer:

Given: Radii of concentric circles: r = 5 cm, R = 13 cm.

Step 1: Draw the diagram showing the two circles and the chord AB of the larger circle touching the smaller circle at point P.

Step 2: Since AB is a tangent to the smaller circle, OP is perpendicular to AB (where O is the common center).
Thus, ΔOPA is a right-angled triangle.

Step 3: Apply the Pythagorean theorem in ΔOPA:
OA² = OP² + AP²
13² = 5² + AP²
169 = 25 + AP²
AP² = 144 ⇒ AP = 12 cm.

Step 4: Since OP bisects AB (perpendicular from center to chord), AB = 2 × AP = 24 cm.

Final Answer: The length of the chord is 24 centimeters.

Question 13:

In a park, a circular fountain is surrounded by a path of uniform width. The radius of the fountain is 7 meters, and the total area of the fountain along with the path is 154 square meters. A child runs around the outer edge of the path. Calculate the distance covered by the child in one complete round.

Answer:

Given, radius of fountain (r) = 7 meters.
Total area of fountain + path = 154 m².
Let the width of the path be w meters.

Step 1: Calculate area of fountain.
Area = πr² = (22/7) × 7 × 7 = 154 m².

Step 2: Total area (fountain + path) = π(r + w)² = 154 m².
But the fountain's area alone is 154 m², which implies the path's area is 0.
This suggests the 'path' has negligible width, so the child runs along the fountain's edge.

Step 3: Distance covered = Circumference of fountain.
Circumference = 2πr = 2 × (22/7) × 7 = 44 meters.

Note: The question seems to have inconsistent data (path width = 0), but the logical interpretation leads to the answer.

Question 14:

Two concentric circles have radii 5 cm and 13 cm. A chord AB of the larger circle touches the smaller circle at point P. Find the length of AB.

Answer:

Given, radii of concentric circles: r = 5 cm, R = 13 cm.
Chord AB of the larger circle touches the smaller circle at P.

Step 1: Draw the perpendicular from the center O to chord AB.
Since OP is perpendicular to AB, it bisects AB (property of circles).

Step 2: Apply Pythagoras theorem in ΔOPA.
OA² = OP² + AP²
13² = 5² + AP²
169 = 25 + AP²
AP² = 144
AP = 12 cm.

Step 3: Length of chord AB = 2 × AP = 24 cm.

Key concept: The perpendicular from the center to a chord bisects it, and the relationship can be solved using right triangles.

Question 15:

In a park, a circular fountain is surrounded by a pathway of uniform width. The radius of the fountain is 7 meters, and the area of the pathway is 44π square meters. A student needs to find the width of the pathway. Help them solve the problem step-by-step.

Answer:

To find the width of the pathway, follow these steps:

Step 1: Let the width of the pathway be x meters.

Step 2: The radius of the fountain (inner circle) = 7 meters.

Step 3: The radius of the outer circle (fountain + pathway) = (7 + x) meters.

Step 4: Area of the pathway = Area of outer circle - Area of inner circle.

Given, area of pathway = 44π.

So, π(7 + x)² - π(7)² = 44π.

Step 5: Simplify the equation:

(7 + x)² - 49 = 44.

49 + 14x + x² - 49 = 44.

14x + x² = 44.

Step 6: Rearrange and solve the quadratic equation:

x² + 14x - 44 = 0.

Using the quadratic formula:

x = [-14 ± √(196 + 176)] / 2.

x = [-14 ± √372] / 2.

Since width cannot be negative, x ≈ 2.6 meters (approximate).

Conclusion: The width of the pathway is approximately 2.6 meters.

Question 16:

Two circles with radii 5 cm and 3 cm intersect at two points, and the distance between their centers is 4 cm. A student is asked to find the length of the common chord. Guide them through the solution.

Answer:

To find the length of the common chord, follow these steps:

Step 1: Draw the two circles with centers O (radius 5 cm) and O' (radius 3 cm), intersecting at points A and B.

Step 2: The distance between centers OO' = 4 cm.

Step 3: Let the common chord AB intersect OO' at point P.

Step 4: Since AB is perpendicular to OO', triangles OPA and O'PA are right-angled.

Step 5: Let OP = x, then O'P = (4 - x).

Step 6: Apply Pythagoras' theorem in both triangles:

In ΔOPA: OA² = OP² + AP² ⇒ 25 = x² + AP².

In ΔO'PA: O'A² = O'P² + AP² ⇒ 9 = (4 - x)² + AP².

Step 7: Subtract the second equation from the first:

25 - 9 = x² - (4 - x)².

16 = x² - (16 - 8x + x²).

16 = 8x - 16.

8x = 32 ⇒ x = 4 cm.

Step 8: Substitute x = 4 into the first equation:

25 = 16 + AP² ⇒ AP² = 9 ⇒ AP = 3 cm.

Step 9: The length of the common chord AB = 2 × AP = 6 cm.

Conclusion: The length of the common chord is 6 cm.

Question 17:

In a park, a circular fountain is surrounded by a pathway of uniform width. The radius of the fountain is 7 meters, and the area of the pathway is 44 square meters. A student needs to find the width of the pathway. Help them solve this step-by-step.

Answer:

To find the width of the pathway, follow these steps:

Step 1: Let the width of the pathway be 'x' meters.
Step 2: The radius of the fountain (inner circle) = 7 meters.
Step 3: The radius of the outer circle (fountain + pathway) = (7 + x) meters.
Step 4: Area of the pathway = Area of outer circle - Area of inner circle.
Step 5: Given, area of pathway = 44 m².
Step 6: So, π(7 + x)² - π(7)² = 44.
Step 7: Simplify: π[(7 + x)² - 49] = 44.
Step 8: Take π = 22/7: (22/7)[(49 + 14x + x²) - 49] = 44.
Step 9: Further simplify: (22/7)(14x + x²) = 44.
Step 10: Multiply both sides by 7/22: 14x + x² = 14.
Step 11: Rearrange: x² + 14x - 14 = 0.
Step 12: Solve the quadratic equation using the formula:
x = [-14 ± √(196 + 56)] / 2 = [-14 ± √252]/2.
Step 13: Simplify √252 = √(36 × 7) = 6√7.
Step 14: So, x = [-14 ± 6√7]/2 = -7 ± 3√7.
Step 15: Since width cannot be negative, x = -7 + 3√7 ≈ 1.93 meters.

Thus, the width of the pathway is approximately 1.93 meters.

Question 18:

Two circles with radii 5 cm and 3 cm intersect at two points, and the distance between their centers is 4 cm. A student needs to find the length of the common chord. Explain the solution clearly.

Answer:

To find the length of the common chord, follow these steps:

Step 1: Let the two circles have centers O (radius = 5 cm) and O' (radius = 3 cm).
Step 2: The distance between centers OO' = 4 cm.
Step 3: Let the common chord be AB, intersecting OO' at M.
Step 4: Since AB is a common chord, OO' is the perpendicular bisector of AB.
Step 5: So, AM = MB, and ∠OMA = 90°.
Step 6: In right ΔOMA, by Pythagoras theorem:
OA² = OM² + AM² ⇒ 25 = OM² + AM².
Step 7: In right ΔO'MA, by Pythagoras theorem:
O'A² = O'M² + AM² ⇒ 9 = (4 - OM)² + AM².
Step 8: Subtract the two equations: 25 - 9 = OM² - (4 - OM)².
Step 9: Simplify: 16 = OM² - (16 - 8OM + OM²).
Step 10: Further simplify: 16 = 8OM - 16 ⇒ 8OM = 32 ⇒ OM = 4 cm.
Step 11: Substitute OM = 4 cm into the first equation: 25 = 16 + AM² ⇒ AM² = 9 ⇒ AM = 3 cm.
Step 12: Since AB = 2 × AM, the length of the common chord AB = 6 cm.

Thus, the length of the common chord is 6 cm.

Question 19:

In a park, a circular fountain is surrounded by a pathway of uniform width. The radius of the fountain is 7 meters, and the total area of the fountain along with the pathway is 154 square meters. A gardener wants to plant flowers along the outer edge of the pathway.

Based on this information, answer the following:

Find the width of the pathway.
Calculate the length of the outer boundary of the pathway (use π = 22/7).

Answer:

Step 1: Find the width of the pathway.

Let the width of the pathway be x meters.
Total radius (fountain + pathway) = 7 + x meters.
Total area = π(7 + x)² = 154 m².
Substitute π = 22/7:
(22/7)(7 + x)² = 154
(7 + x)² = 154 × (7/22) = 49
7 + x = √49 = 7
x = 0.

This suggests the pathway has zero width, meaning the given total area matches the fountain's area (πr² = 154 m²). There might be an error in the problem statement.

Step 2: Length of outer boundary (if pathway existed).

If pathway width were x, outer boundary = 2π(7 + x).
But since x = 0, outer boundary = 2π(7) = 44 meters.

Note: The question likely intended a larger total area to yield a non-zero pathway width. Always verify given values for consistency.

Question 20:

Two concentric circles have radii 5 cm and 13 cm. A chord AB of the larger circle touches the smaller circle at point P.

Answer the following:

Prove that AP = BP.
Find the length of chord AB.

Answer:

Proof that AP = BP:

1. Draw the perpendicular from the center O to chord AB.
2. Since AB touches the smaller circle at P, OP is the radius of the smaller circle and perpendicular to AB.
3. Perpendicular from center bisects the chord: AP = BP (Proved).

Length of chord AB:

1. In ΔOAP, OA = 13 cm (radius of larger circle), OP = 5 cm (radius of smaller circle).
2. By Pythagoras' theorem:
AP = √(OA² - OP²) = √(169 - 25) = √144 = 12 cm.
3. Since AP = BP, AB = 2 × AP = 24 cm.

Key concept: The tangent to a circle is perpendicular to the radius at the point of contact, enabling right-angle triangle properties.

Circles – CBSE NCERT Study Resources

Overview of the Chapter: Circles

Key Topics Covered

Tangent to a Circle

Number of Tangents from a Point to a Circle

Lengths of Tangents

Theorems Related to Chords and Angles

All Question Types with Solutions – CBSE Exam Pattern

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)