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Circles – CBSE NCERT Study Resources

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9th

9th - Mathematics

Circles

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Chapter Overview: Circles

This chapter introduces the fundamental concepts related to circles, including their properties, terms, and theorems. Students will learn about chords, arcs, angles subtended by chords, and cyclic quadrilaterals as per the CBSE Grade 9 Mathematics syllabus.

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 Concepts

  • Circle and its related terms: radius, diameter, chord, arc, segment, sector, etc.
  • Angle subtended by a chord at a point
  • Perpendicular from the center to a chord
  • Circle through three points
  • Equal chords and their distances from the center
  • Angle subtended by an arc of a circle
  • Cyclic quadrilaterals

Important Theorems

Theorem 1: Equal chords of a circle subtend equal angles at the center.

Theorem 2: The perpendicular from the center of a circle to a chord bisects the chord.

Theorem 3: There is one and only one circle passing through three non-collinear points.

Theorem 4: Equal chords of a circle are equidistant from the center.

Theorem 5: The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Theorem 6: Angles in the same segment of a circle are equal.

Theorem 7: If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, the four points lie on a circle.

Theorem 8: The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

Applications

Understanding these concepts helps in solving geometric problems related to circles, including finding lengths, angles, and proving various properties in different geometric configurations.

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:
Define the radius of a circle.
Answer:

The distance from the center to any point on the circle.

Question 2:
What is the diameter of a circle with radius 7 cm?
Answer:
14 cm
Question 3:
Name the longest chord in a circle.
Answer:

Diameter is the longest chord.

Question 4:
If the circumference of a circle is 44 cm, find its radius. (Use π = 22/7)
Answer:
7 cm
Question 5:
What is the measure of a semicircular arc?
Answer:
180°
Question 6:
How many tangents can be drawn from an external point to a circle?
Answer:
2
Question 7:
If two circles touch externally, what is the distance between their centers?
Answer:

Sum of their radii.

Question 8:
What is the angle subtended by a diameter in a semicircle?
Answer:
90°
Question 9:
Find the area of a circle with radius 3.5 cm. (Use π = 22/7)
Answer:
38.5 cm²
Question 10:
What is the circumference of a circle with diameter 10 cm? (Use π = 3.14)
Answer:
31.4 cm
Question 11:
If two circles are congruent, what must be equal?
Answer:

Their radii must be equal.

Question 12:
What is the length of an arc subtending 60° in a circle of radius 6 cm? (Use π = 22/7)
Answer:
6.28 cm
Question 13:
Define the term radius of a circle.
Answer:

The radius of a circle is the distance from the center of the circle to any point on its circumference.

Question 14:
What is the longest chord of a circle called?
Answer:

The longest chord of a circle is called the diameter. It passes through the center and is twice the length of the radius.

Question 15:
If the diameter of a circle is 14 cm, what is its radius?
Answer:

Given: Diameter = 14 cm
Radius = Diameter / 2
Radius = 14 cm / 2 = 7 cm.

Question 16:
What is the measure of a semicircular arc in degrees?
Answer:

A semicircular arc measures 180° because a full circle is 360° and a semicircle is half of it.

Question 17:
State the relationship between the radius and diameter of a circle.
Answer:

The diameter of a circle is twice the length of its radius.
Mathematically, Diameter = 2 × Radius.

Question 18:
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 a point outside it.

Question 19:
What is the angle between the radius and the tangent at the point of contact?
Answer:

The angle between the radius and the tangent at the point of contact is always 90°.

Question 20:
If two circles intersect at two points, what is the maximum number of common tangents they can have?
Answer:

Two circles intersecting at two points can have a maximum of two common tangents.

Question 21:
What is the perimeter of a semicircle with radius 7 cm? (Use π = 22/7)
Answer:

Perimeter of semicircle = πr + 2r
= (22/7 × 7) + (2 × 7)
= 22 + 14 = 36 cm.

Question 22:
Define the term cyclic quadrilateral.
Answer:

A cyclic quadrilateral is a four-sided figure where all the vertices lie on the circumference of a single circle.

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 chord of a circle.
Answer:

A chord is a straight line segment whose endpoints lie on the circumference of the circle.
It is different from the diameter as it does not necessarily pass through the center.

Question 2:
What is the relationship between the radius and diameter of a circle?
Answer:

The diameter of a circle is twice the length of its radius.
Mathematically, Diameter = 2 × Radius.

Question 3:
State the perpendicular from the center to a chord theorem.
Answer:

The perpendicular from the center of a circle to a chord bisects the chord.
This means it divides the chord into two equal parts.

Question 4:
If the length of a chord is 10 cm and the radius is 13 cm, find the distance of the chord from the center.
Answer:

Using the formula: Distance (d) = √(r² - (l/2)²)
Here, r = 13 cm, l = 10 cm
d = √(13² - 5²)
d = √(169 - 25)
d = √144
d = 12 cm

Question 5:
What is the measure of the angle subtended by the diameter at any point on the circle?
Answer:

The angle subtended by the diameter at any point on the circle is a right angle (90°).
This is known as the Thales' theorem.

Question 6:
How many circles can pass through three non-collinear points?
Answer:

Only one unique circle can pass through three non-collinear points.
This is because three points define a unique circumcircle.

Question 7:
If two chords of a circle are equal in length, what can you say about their distances from the center?
Answer:

Equal chords of a circle are equidistant from the center.
This is a direct result of the chord length theorem.

Question 8:
What is the cyclic quadrilateral? State one property.
Answer:

A cyclic quadrilateral is a four-sided figure where all vertices lie on the circumference of a circle.
Property: The sum of opposite angles is 180° (supplementary).

Question 9:
Find the length of a chord which is at a distance of 5 cm from the center of a circle with radius 13 cm.
Answer:

Using the formula: Chord length (l) = 2√(r² - d²)
Here, r = 13 cm, d = 5 cm
l = 2√(13² - 5²)
l = 2√(169 - 25)
l = 2√144
l = 2 × 12 = 24 cm

Question 10:
What is the arc of a circle? How is it related to the central angle?
Answer:

An arc is a portion of the circumference of the circle.
The length of the arc is directly proportional to the central angle subtended by it.
For a full circle (360°), the arc length is the circumference.

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:
Define a circle and explain the relationship between its radius and diameter.
Answer:

A circle is a closed two-dimensional shape where all points on its boundary are equidistant from a fixed point called the center.
The radius is the distance from the center to any point on the circle.
The diameter is twice the radius, i.e., it passes through the center and connects two points on the boundary.
Mathematically, Diameter (D) = 2 × Radius (r).

Question 2:
Prove that the perpendicular from the center of a circle to a chord bisects the chord.
Answer:

Consider a circle with center O and chord AB.
Draw a perpendicular OM from O to AB.
In triangles OAM and OBM:
OA = OB (radii of the same circle)
OM is common
∠OMA = ∠OMB = 90° (perpendicular)
Thus, ΔOAM ≅ ΔOBM by RHS congruency.
Therefore, AM = BM, proving the perpendicular bisects the chord.

Question 3:
Explain why equal chords of a circle are equidistant from the center.
Answer:

Let AB and CD be two equal chords in a circle with center O.
Draw perpendiculars OM and ON from O to AB and CD, respectively.
Since AB = CD, their perpendicular distances from the center are equal (OM = ON).
This is because in congruent right triangles OAM and OCN:
OA = OC (radii)
AM = CN (half of equal chords)
Thus, by Pythagoras theorem, OM = ON.

Question 4:
If the angle subtended by a chord at the center is 60°, what is the angle subtended by the same chord at any point on the remaining part of the circle?
Answer:

According to the Central Angle Theorem, the angle subtended by a chord at the center is twice the angle subtended at any point on the remaining part of the circle.
Given, central angle = 60°.
Thus, angle at the circumference = 60° / 2 = 30°.
This is a direct application of the theorem and holds true for all circles.

Question 5:
Describe how to construct a tangent to a circle from an external point using a compass and ruler.
Answer:

Steps to construct a tangent:
1. Draw a circle with center O and mark an external point P.
2. Join OP and find its midpoint M.
3. With M as center and MO as radius, draw a semicircle intersecting the circle at T.
4. Join PT, which is the required tangent.

This works because ∠OTP is 90° (angle in a semicircle), making PT perpendicular to OT, the radius at the point of contact.

Question 6:
Prove that equal chords of a circle subtend equal angles at the center.
Answer:

Consider a circle with center O and two equal chords AB and CD.
Join OA, OB, OC, and OD.
In triangles OAB and OCD:
1. OA = OC (radii of the same circle)
2. OB = OD (radii of the same circle)
3. AB = CD (given, equal chords)
Thus, by SSS congruence, ∆OAB ≅ ∆OCD.
Therefore, ∠AOB = ∠COD (corresponding angles of congruent triangles).

Question 7:
If the length of a chord of a circle is 16 cm and its perpendicular distance from the center is 6 cm, find the radius of the circle.
Answer:

Given:
Length of chord (AB) = 16 cm
Perpendicular distance from center (OM) = 6 cm

Using the formula:
Radius (r) = √[(AB/2)² + OM²]
Step 1: Half of chord length = 16/2 = 8 cm
Step 2: Apply Pythagoras theorem:
r = √(8² + 6²) = √(64 + 36) = √100 = 10 cm
Thus, the radius of the circle is 10 cm.

Question 8:
Explain why a cyclic quadrilateral has its opposite angles supplementary.
Answer:

A cyclic quadrilateral is a four-sided figure inscribed in a circle, with all vertices lying on the circumference.
By the Inscribed Angle Theorem, the opposite angles of a cyclic quadrilateral are supplementary because:
1. Each angle is half the measure of the arc it subtends.
2. The sum of the arcs subtended by opposite angles covers the entire circle (360°).
Thus, if one pair of opposite angles subtends arcs totaling 180°, their sum is 180° (supplementary).

Question 9:
Draw a circle with radius 5 cm. Construct two tangents from an external point 7 cm away from the center.
Answer:

Steps to construct tangents:
1. Draw a circle with center O and radius 5 cm.
2. Mark an external point P, 7 cm from O.
3. Join OP and draw its perpendicular bisector to locate midpoint M.
4. With M as center and MO as radius, draw a semicircle intersecting the circle at T₁ and T₂.
5. Join PT₁ and PT₂, which are the required tangents.
Note: The lengths of both tangents will be equal due to the Tangent Theorem.

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 equal chords of a circle subtend equal angles at the center. Use a diagram and step notation.
Answer:
Introduction

We studied that chords and angles in a circle are related. Our textbook shows that equal chords create equal angles at the center.


Argument 1
  • Draw a circle with center O and two equal chords AB and CD.
  • Join OA, OB, OC, and OD.

Argument 2
  • Triangles OAB and OCD are congruent by SSS (OA=OC, OB=OD, AB=CD).
  • Thus, ∠AOB = ∠COD.

Conclusion

This proves equal chords subtend equal angles at the center.

Question 2:
A bicycle wheel has a radius of 35 cm. Calculate its circumference and explain one real-life application of this concept.
Answer:
Introduction

We know the circumference of a circle is calculated using C = 2πr. Our textbook uses this formula for practical problems.


Argument 1
  • Given radius (r) = 35 cm.
  • C = 2 × (22/7) × 35 = 220 cm.

Argument 2

In real life, this helps in designing wheels for vehicles, ensuring smooth motion.


Conclusion

The circumference is 220 cm, useful in engineering and design.

Question 3:
Explain why perpendicular from the center to a chord bisects it. Derive this property with a diagram.
Answer:
Introduction

Our textbook states that a perpendicular from the center to a chord divides it equally.


Argument 1
  • Draw a circle with center O, chord AB, and perpendicular OM.
  • Triangles OMA and OMB are congruent by RHS (OM common, OA=OB as radii).

Argument 2
  • Thus, AM = MB, proving the chord is bisected.

Conclusion

The perpendicular from the center bisects the chord, a key property in circle geometry.

Question 4:
Prove that equal chords of a circle subtend equal angles at the center. Support your answer with a diagram and an example from NCERT.
Answer:
Introduction

We studied that chords are line segments joining two points on a circle. Equal chords have special properties.


Argument 1

Consider two equal chords AB and CD of a circle with center O. In ΔAOB and ΔCOD, AB = CD (given), OA = OC (radii), and OB = OD (radii). By SSS congruence, ΔAOB ≅ ΔCOD.


Argument 2

Thus, ∠AOB = ∠COD (CPCT). Our textbook shows this in Example 1 of Chapter 10.


Conclusion

Hence, equal chords subtend equal angles at the center. [Diagram: Two equal chords AB and CD subtending ∠AOB and ∠COD at center O.]

Question 5:
A bicycle wheel has a radius of 35 cm. Calculate the number of complete revolutions it makes to cover a distance of 1.1 km. Use π = 22/7.
Answer:
Introduction

We know the circumference of a circle gives the distance covered in one revolution.


Argument 1

Circumference = 2πr = 2 × (22/7) × 35 cm = 220 cm. Total distance = 1.1 km = 110,000 cm.


Argument 2

Number of revolutions = Total distance / Circumference = 110,000 cm / 220 cm = 500. This matches NCERT Exercise 12.1 Problem 5.


Conclusion

The wheel completes 500 revolutions to cover 1.1 km. Real-life applications include odometer calculations.

Question 6:
Explain why three non-collinear points always lie on a unique circle. Derive the steps to find its center.
Answer:
Introduction

Three non-collinear points form a triangle, and we can draw a circle passing through them.


Argument 1

Let points A, B, and C be non-collinear. Draw perpendicular bisectors of AB and BC. They intersect at O, the center.


Argument 2

Since OA = OB = OC (radii), the circle is unique. Our textbook derives this in Theorem 10.5.


Conclusion

Thus, three non-collinear points lie on one unique circle. [Diagram: Triangle ABC with perpendicular bisectors meeting at O.]

Question 7:
Prove that equal chords of a circle subtend equal angles at the center. Use a diagram and step-wise reasoning.
Answer:
Introduction

We studied that chords and angles in a circle are related. Our textbook shows that equal chords create equal angles at the center.


Argument 1
  • Draw a circle with center O and two equal chords AB and CD.
  • Join OA, OB, OC, OD to form ΔOAB and ΔOCD.

Argument 2
  • Since AB = CD (given), OA = OB = OC = OD (radii).
  • By SSS congruency, ΔOAB ≅ ΔOCD, so ∠AOB = ∠COD.

Conclusion

Thus, equal chords subtend equal angles at the center, as proved.

Question 8:
A bicycle wheel has a radius of 35 cm. Calculate its circumference and explain how this applies in real-life measurements.
Answer:
Introduction

We learned that circumference is the distance around a circle. Our textbook uses the formula C = 2πr.


Argument 1
  • Given radius (r) = 35 cm.
  • Circumference (C) = 2 × (22/7) × 35 = 220 cm.

Argument 2

In real life, this helps measure the distance covered in one wheel rotation. For example, a bicycle covers 220 cm per revolution.


Conclusion

Thus, the wheel’s circumference is 220 cm, useful for motion calculations.

Question 9:
Two circles of radii 5 cm and 3 cm intersect at two points. Prove that the perpendicular bisector of their common chord passes through their centers.
Answer:
Introduction

We studied that the common chord of two intersecting circles has special properties. Our textbook shows the perpendicular bisector’s role.


Argument 1
  • Let circles with centers O (5 cm) and O' (3 cm) intersect at A and B.
  • AB is the common chord.

Argument 2
  • The perpendicular bisector of AB passes through O and O' as equal chords (from centers) subtend equal angles.
  • Thus, OO' is the perpendicular bisector of AB.

Conclusion

Hence, the perpendicular bisector of AB passes through both centers.

Question 10:
Prove that equal chords of a circle subtend equal angles at the center. Use a diagram and step-by-step reasoning.
Answer:
Introduction

We studied that chords are line segments joining two points on a circle. Our textbook shows that equal chords have special properties.


Argument 1
  • Draw a circle with center O and two equal chords AB and CD.
  • Join OA, OB, OC, and OD to form triangles OAB and OCD.

Argument 2
  • Since AB = CD (given), and OA = OB = OC = OD (radii), triangles OAB and OCD are congruent by SSS rule.
  • Thus, ∠AOB = ∠COD (CPCT).

Conclusion

This proves equal chords subtend equal angles at the center, a key concept in circle geometry.

Question 11:
A bicycle wheel has a radius of 35 cm. Calculate its circumference and explain how this applies in real-life situations.
Answer:
Introduction

We learned that circumference is the distance around a circle. Our textbook uses the formula C = 2πr.


Argument 1
  • Given radius (r) = 35 cm, circumference C = 2 × (22/7) × 35 = 220 cm.
  • This means one full wheel rotation covers 220 cm.

Argument 2
  • In real life, this helps calculate distance traveled: if the wheel rotates 100 times, distance = 220 × 100 = 22,000 cm (220 m).
  • Bicycle odometers use this principle.

Conclusion

Understanding circumference has practical applications in designing wheels and measuring distances.

Question 12:
Construct a tangent to a circle from a point outside it, justifying each step with reasoning.
Answer:
Introduction

We studied that a tangent touches a circle at exactly one point. Our textbook shows how to construct it.


Argument 1
  • Draw circle with center O and external point P. Join OP.
  • Find midpoint M of OP and draw a circle with diameter OP.

Argument 2
  • The two circles intersect at T. Join PT, which is the required tangent.
  • Justification: ∠OTP = 90° (angle in semicircle), so PT ⊥ OT.

Conclusion

This method ensures PT touches the circle only at T, satisfying the tangent definition.

Question 13:
Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Provide a step-by-step proof with a diagram.
Answer:

To prove that the angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle, follow these steps:


Given: A circle with centre O and an arc AB subtending ∠AOB at the centre and ∠ACB at any point C on the remaining part of the circle.
To Prove: ∠AOB = 2 × ∠ACB

Construction: Join CO and extend it to point D.

Proof:
1. In △AOC, OA = OC (radii of the same circle).
∴ △AOC is an isosceles triangle.
⇒ ∠OAC = ∠OCA (Angles opposite equal sides are equal).
Let ∠OAC = ∠OCA = x.
By exterior angle property, ∠AOD = ∠OAC + ∠OCA = x + x = 2x.

2. Similarly, in △BOC, OB = OC (radii of the same circle).
∴ △BOC is an isosceles triangle.
⇒ ∠OBC = ∠OCB (Angles opposite equal sides are equal).
Let ∠OBC = ∠OCB = y.
By exterior angle property, ∠BOD = ∠OBC + ∠OCB = y + y = 2y.

3. Now, ∠AOB = ∠AOD + ∠BOD = 2x + 2y = 2(x + y).
But ∠ACB = ∠OCA + ∠OCB = x + y.
∴ ∠AOB = 2 × ∠ACB (Proved).

Diagram: Draw a circle with centre O, points A, B, C on the circumference, and D outside the circle on the extension of CO.
Question 14:
Explain why opposite angles of a cyclic quadrilateral are supplementary. Support your answer with a diagram and step-by-step reasoning.
Answer:

A cyclic quadrilateral is a quadrilateral whose all four vertices lie on a single circle. The property that opposite angles of a cyclic quadrilateral are supplementary (i.e., they add up to 180°) can be proven as follows:


Given: A cyclic quadrilateral ABCD inscribed in a circle with centre O.
To Prove: ∠A + ∠C = 180° and ∠B + ∠D = 180°.

Proof:
1. Consider the quadrilateral ABCD with vertices on the circle.
Let arc BCD subtend ∠A at the remaining part of the circle.
Let arc BAD subtend ∠C at the remaining part of the circle.

2. By the angle subtended by an arc property, the angle subtended by arc BCD at the centre is 2 × ∠A.
Similarly, the angle subtended by arc BAD at the centre is 2 × ∠C.

3. The sum of angles subtended by arcs BCD and BAD at the centre is 360° (complete circle).
∴ 2∠A + 2∠C = 360°.
⇒ ∠A + ∠C = 180°.

4. Similarly, by considering arcs ABC and ADC, we can prove ∠B + ∠D = 180°.

Diagram: Draw a circle with a quadrilateral ABCD inscribed in it, showing arcs and angles as described.
Question 15:
Describe how to construct a tangent to a circle from an external point. Provide a detailed step-by-step procedure with a diagram.
Answer:

To construct a tangent to a circle from an external point, follow these steps:


Given: A circle with centre O and an external point P.
To Construct: Tangents from P to the circle.

Steps:
1. Join OP.
2. Draw the perpendicular bisector of OP to find its midpoint M.
3. With M as the centre and MO as the radius, draw a circle intersecting the given circle at points T and T'.
4. Join PT and PT'. These are the required tangents.

Justification:
1. Since ∠OTP is an angle in a semicircle, it is a right angle (90°).
2. Therefore, OT ⊥ PT, making PT a tangent (as a tangent is perpendicular to the radius at the point of contact).
3. Similarly, PT' is also a tangent.

Diagram: Draw the given circle with centre O, point P outside it, and the construction steps as described, showing PT and PT' as tangents.
Question 16:
Prove that equal chords of a circle subtend equal angles at the centre. Support your answer with a diagram.
Answer:

To prove that equal chords of a circle subtend equal angles at the centre, let us consider a circle with centre O and two equal chords AB and CD.

Step 1: Draw the diagram showing the circle with centre O, chords AB and CD, and the radii OA, OB, OC, and OD.

Step 2: Since AB = CD (given), and all radii of the same circle are equal (OA = OB = OC = OD), the triangles OAB and OCD are congruent by the SSS (Side-Side-Side) rule.

Step 3: In congruent triangles, corresponding angles are equal. Hence, ∠AOB = ∠COD.

Conclusion: Therefore, equal chords subtend equal angles at the centre of the circle.

Question 17:
Explain why the perpendicular from the centre of a circle to a chord bisects the chord. Illustrate with a diagram.
Answer:

The perpendicular from the centre of a circle to a chord bisects the chord because of the following reasoning:

Step 1: Draw a circle with centre O and a chord AB. Draw a perpendicular OM from O to AB, meeting it at point M.

Step 2: In triangles OAM and OBM, OA = OB (radii of the same circle), OM is common, and ∠OMA = ∠OMB = 90°.

Step 3: By the RHS (Right Angle-Hypotenuse-Side) congruence rule, triangles OAM and OBM are congruent.

Step 4: Therefore, AM = MB, proving that the perpendicular bisects the chord.

Conclusion: The perpendicular from the centre to the chord splits it into two equal parts.

Question 18:
Describe how to find the length of a chord when the distance from the centre of the circle to the chord is given. Use an example with a circle of radius 5 cm and a distance of 3 cm from the centre to the chord.
Answer:

To find the length of a chord when the distance from the centre to the chord is known, follow these steps:

Step 1: Recall the relationship between the radius (r), the perpendicular distance (d) from the centre to the chord, and the length of the chord (L): L = 2√(r² - d²).

Step 2: Given r = 5 cm and d = 3 cm, substitute these values into the formula.

Step 3: Calculate L = 2√(5² - 3²) = 2√(25 - 9) = 2√16 = 2 × 4 = 8 cm.

Conclusion: The length of the chord is 8 cm. This method is derived from the Pythagorean theorem applied to the right triangle formed by the radius, the perpendicular, and half the chord.

Question 19:
Prove that the lengths of tangents drawn from an external point to a circle are equal. Support your answer with a diagram.
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.
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 RHS congruence rule, ΔOAP ≅ ΔOBP.
3. Therefore, by CPCT, PA = PB.

Hence, 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. Connect OA, OB, and OP.)
Question 20:
In a circle of radius 5 cm, a chord subtends a right angle at the center. Find the length of the chord. Show your working clearly.
Answer:

To find the length of the chord subtending a right angle at the center of a circle with radius 5 cm, follow these steps:


Given:
- Radius (r) of the circle = 5 cm
- Angle subtended by the chord at the center (θ) = 90°
To Find: Length of the chord (AB)
Solution:
1. The formula to find the length of a chord when the angle subtended at the center is known is:
Length of chord = 2 × r × sin(θ/2)
2. Substitute the given values:
Length of chord = 2 × 5 × sin(90°/2)
Length of chord = 10 × sin(45°)
3. We know that sin(45°) = 1/√2, so:
Length of chord = 10 × (1/√2) = 10/√2
4. Rationalizing the denominator:
Length of chord = (10 × √2) / (√2 × √2) = (10√2)/2 = 5√2 cm

Thus, the length of the chord is 5√2 cm.


Diagram: (Draw a circle with center O, radius 5 cm, and a chord AB subtending a 90° angle at O.)
Question 21:
Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. (5 marks)
Answer:

To prove that the angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle, we follow these steps:



Given: A circle with centre O and an arc AB subtending ∠AOB at the centre and ∠ACB at any point C on the remaining part of the circle.

To Prove: ∠AOB = 2 × ∠ACB

Construction: Join CO and extend it to a point D.

Proof:

1. In △AOC, OA = OC (radii of the same circle).
∴ △AOC is an isosceles triangle.
⇒ ∠OAC = ∠OCA (Angles opposite to equal sides are equal).

2. By the Exterior Angle Theorem, ∠AOD = ∠OAC + ∠OCA.
⇒ ∠AOD = 2 × ∠OCA (since ∠OAC = ∠OCA).

3. Similarly, in △BOC, OB = OC (radii of the same circle).
∴ △BOC is an isosceles triangle.
⇒ ∠OBC = ∠OCB.

4. By the Exterior Angle Theorem, ∠BOD = ∠OBC + ∠OCB.
⇒ ∠BOD = 2 × ∠OCB (since ∠OBC = ∠OCB).

5. Now, ∠AOB = ∠AOD + ∠BOD.
⇒ ∠AOB = 2 × ∠OCA + 2 × ∠OCB.
⇒ ∠AOB = 2 × (∠OCA + ∠OCB).
⇒ ∠AOB = 2 × ∠ACB (since ∠ACB = ∠OCA + ∠OCB).

Hence, the angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle.

Question 22:
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's follow these 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. Since PA is a tangent at A, OA is perpendicular to PA (Radius ⊥ Tangent).
∴ ∠OAP = 90°.
2. Similarly, OB is perpendicular to PB.
∴ ∠OBP = 90°.
3. In right triangles OAP and OBP:
- OA = OB (Radii of the same circle).
- OP = OP (Common side).
4. By the RHS Congruence Rule, ΔOAP ≅ ΔOBP.
5. Therefore, by CPCT, 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 tangents PA and PB touching the circle at A and B respectively. Show radii OA and OB perpendicular to the tangents.)

Value-Added Insight: This property is useful in solving real-world problems, such as finding the distance between two points of contact when a tangent is drawn from a fixed external point to a circular object.

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 has a radius of 14 meters. A gardener wants to fence it with wire. Calculate the length of wire required. Also, find the cost if the wire costs ₹50 per meter.
Answer:
Problem Interpretation

We need to find the circumference of the park (length of wire) and then calculate the cost.


Mathematical Modeling

Circumference (C) = 2πr, where r = 14m.


Solution
  • C = 2 × (22/7) × 14 = 88m
  • Cost = 88 × 50 = ₹4400
Question 2:
In a circle with center O, chord AB is 10 cm long and the perpendicular distance from O to AB is 12 cm. Find the radius of the circle.
Answer:
Problem Interpretation

We studied that the perpendicular from the center bisects the chord. Here, AB = 10 cm, so half of AB is 5 cm.


Mathematical Modeling

Using Pythagoras theorem: r² = (distance)² + (half chord)².


Solution
  • r² = 12² + 5² = 144 + 25 = 169
  • r = √169 = 13 cm
Question 3:
A circular park has a radius of 14 m. A path of uniform width 7 m is built around it. Find the area of the path. (Use π = 22/7)
Answer:
Problem Interpretation

We need to find the area of the path around the park, which forms a ring.


Mathematical Modeling
  • Inner radius (park) = 14 m
  • Outer radius (park + path) = 14 + 7 = 21 m

Solution

Area of path = π(R² - r²) = (22/7)(21² - 14²) = (22/7)(441 - 196) = (22/7)(245) = 770 m².

Question 4:
Two concentric circles have radii 5 cm and 13 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Problem Interpretation

We studied that the chord of the larger circle is tangent to the smaller circle.


Mathematical Modeling
  • Let AB be the chord. Draw perpendicular OP from center O to AB.
  • OP = radius of smaller circle = 5 cm.

Solution

Using Pythagoras’ theorem in ΔOPA, AP = √(13² - 5²) = √144 = 12 cm. Thus, AB = 2 × 12 = 24 cm.

Question 5:
A circular park has a radius of 14 m. A gardener wants to plant flowers along the circumference. If each flower requires 0.5 m of space, how many flowers can be planted?
Answer:
Problem Interpretation

We need to find the number of flowers that fit along the circumference of a circular park.

Mathematical Modeling
  • Radius (r) = 14 m
  • Space per flower = 0.5 m
Solution

Circumference = 2πr = 2 × (22/7) × 14 = 88 m. Number of flowers = Circumference / Space per flower = 88 / 0.5 = 176 flowers.

Question 6:
In a circle with center O, two chords AB and CD are equal in length (6 cm each). If the perpendicular distance from O to AB is 4 cm, what is the distance from O to CD?
Answer:
Problem Interpretation

We studied that equal chords are equidistant from the center. Here, AB = CD = 6 cm.

Mathematical Modeling
  • Length of AB = CD = 6 cm
  • Perpendicular distance (OM) to AB = 4 cm
Solution

Since chords are equal, distance from O to CD = distance from O to AB = 4 cm.

Question 7:
A circular park has a radius of 14 m. A path of uniform width 7 m runs around it. Calculate the area of the path. (Use π = 22/7)
Answer:
Problem Interpretation

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


Mathematical Modeling
  • Inner radius (park) = 14 m
  • Outer radius (park + path) = 14 + 7 = 21 m

Solution

Area of path = π(R² - r²) = (22/7)(21² - 14²) = (22/7)(441 - 196) = (22/7)(245) = 770 m².

Question 8:
Two concentric circles have radii 5 cm and 13 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Problem Interpretation

We studied that a chord of the larger circle touching the smaller circle is tangent to it.


Mathematical Modeling
  • Let AB be the chord. Draw perpendicular OC from center O to AB.
  • OC = radius of smaller circle = 5 cm

Solution

Using Pythagoras theorem in ΔOAC: AC = √(13² - 5²) = √(169 - 25) = 12 cm. Thus, AB = 2 × 12 = 24 cm.

Question 9:
In a circle with center O, chord AB is 12 cm long. The perpendicular distance from O to AB is 8 cm. Find the radius of the circle.
Answer:
Problem Interpretation

We studied that perpendicular from center bisects the chord. Here, AB = 12cm, so AM = 6cm.


Mathematical Modeling
  • Let radius = r
  • Distance OM = 8cm
  • AM = 6cm

Solution

Using Pythagoras theorem: r² = OM² + AM² = 8² + 6² = 64 + 36 = 100. Thus, r = 10cm.

Question 10:
A circular park has a radius of 14 meters. A path of uniform width 7 meters is built around it. Find the area of the path. (Use π = 22/7)
Answer:
Problem Interpretation

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


Mathematical Modeling
  • Inner radius (park) = 14m
  • Outer radius (park + path) = 14m + 7m = 21m

Solution

Area of path = π(R² - r²) = (22/7)(21² - 14²) = (22/7)(441 - 196) = (22/7)(245) = 770m².

Question 11:
A circular park has a radius of 14 meters. A gardener wants to plant flowers along the circumference. Calculate the length of the boundary where flowers will be planted. (Use π = 22/7)
Answer:
Problem Interpretation

We need to find the circumference of the circular park to determine the length where flowers will be planted.


Mathematical Modeling

Circumference (C) of a circle is given by C = 2πr, where r is the radius.


Solution

Given r = 14 m, π = 22/7. Substituting, C = 2 × (22/7) × 14 = 88 m. Thus, the gardener needs to plant flowers along 88 meters.

Question 12:
In a circle with center O, chord AB is 10 cm long, and the perpendicular distance from O to AB is 12 cm. Find the radius of the circle.
Answer:
Problem Interpretation

We studied that the perpendicular from the center to a chord bisects it. Here, we need to find the radius using Pythagoras' theorem.


Mathematical Modeling

Let the radius be r. The perpendicular distance (d) = 12 cm, half of chord AB = 5 cm. Using r² = d² + (AB/2)².


Solution

Substituting, r² = 12² + 5² = 144 + 25 = 169. Thus, r = √169 = 13 cm. The radius is 13 cm.

Question 13:

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 covered by the fountain and the pathway is 154π square meters.

(i) Find the width of the pathway.

(ii) If the pathway is to be paved with tiles costing ₹50 per square meter, calculate the total cost of tiling.

Answer:

Solution:

(i) Let the width of the pathway be w meters.


Radius of fountain (r) = 7 m


Total radius (fountain + pathway) = 7 + w meters


Total area = π(7 + w)² = 154π


Divide both sides by π: (7 + w)² = 154


Expand: 49 + 14w + w² = 154


Simplify: w² + 14w - 105 = 0


Solve using quadratic formula:


w = [-14 ± √(196 + 420)] / 2


w = [-14 ± √616] / 2


w = [-14 ± 24.8] / 2 (taking positive root)


w ≈ 5.4 meters


Pathway width ≈ 5.4 meters


(ii) Area of pathway = Total area - Fountain area


= 154π - π(7)²


= 154π - 49π = 105π ≈ 329.7 m²


Cost = 329.7 × 50 = ₹16,485


Total tiling cost ≈ ₹16,485

Question 14:

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

(i) Prove that AP = PB.

(ii) Find the length of chord AB.

Answer:

Solution:

(i) Proof:


1. Draw perpendicular OP from center O to chord AB of larger circle.


2. Since AB touches smaller circle at P, OP is radius of smaller circle.


3. We know: Perpendicular from center bisects the chord.


4. Thus, AP = PB (Proved).


(ii) Length of AB:


1. In right ΔOPA:


OA = 13 cm (radius of larger circle)


OP = 5 cm (radius of smaller circle)


2. Using Pythagoras theorem:


AP² = OA² - OP² = 169 - 25 = 144


AP = 12 cm


3. Since AP = PB, AB = 2 × AP = 24 cm


Length of chord AB = 24 cm

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 total area covered by the fountain and the pathway is 154π square meters. A child runs around the outer edge of the pathway.

Based on this scenario, answer the following:

1) Calculate the width of the pathway.
2) Find the distance covered by the child in one complete round.

Answer:

1) Width of the pathway:

Let the width of the pathway be 'w' meters.

Radius of fountain (r) = 7 m
Total radius (fountain + pathway) = R = (7 + w) m

Total area = πR² = 154π
=> (7 + w)² = 154
=> 49 + 14w + w² = 154
=> w² + 14w - 105 = 0

Solving the quadratic equation:
w = [-14 ± √(196 + 420)] / 2
w = [-14 ± √616] / 2
w = [-14 ± 24.8] / 2

Taking positive value:
w ≈ 5.4 meters

2) Distance covered by the child:

Outer circumference = 2πR
= 2π(7 + 5.4)
= 2π × 12.4
≈ 77.92 meters

Note: The child covers the circumference of the outer edge, which includes the fountain and pathway.

Question 16:

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

Based on this information, answer the following:

1) Prove that AP = PB.
2) Find the length of chord AB.

Answer:

1) Proof that AP = PB:

Since AB is a chord of the larger circle and touches the smaller circle at P, OP is perpendicular to AB (as radius is perpendicular to tangent at point of contact).

In the larger circle, a perpendicular from the center to a chord bisects it.
Thus, OP ⊥ AB ⇒ AP = PB.

2) Length of chord AB:

Using Pythagoras theorem in ΔOPA:
OA² = OP² + AP²
13² = 5² + AP²
169 = 25 + AP²
AP² = 144
AP = 12 cm

Since AP = PB,
AB = AP + PB = 12 + 12 = 24 cm

Key concept: The perpendicular from the center to a chord bisects it, and the tangent is perpendicular to the radius at the point of contact.

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 total area covered by the fountain and the pathway is 154 square meters. A student claims that the width of the pathway is √2 meters. Verify this claim using mathematical reasoning.

Answer:

To verify the claim, let's calculate the width of the pathway step-by-step.


Given:

  • Radius of fountain (r) = 7 m
  • Total area of fountain + pathway = 154 m²

Step 1: Calculate the area of the fountain.


Area of fountain = πr² = (22/7) × 7 × 7 = 154 m²

Step 2: Let the width of the pathway be x meters.


Total radius (fountain + pathway) = (7 + x) m

Step 3: Total area = π(7 + x)² = 154 m²


(22/7)(7 + x)² = 154
(7 + x)² = 154 × (7/22) = 49
7 + x = √49 = 7
x = 0

Conclusion: The calculation shows the width of the pathway is 0 meters, meaning there is no pathway. The student's claim of √2 meters is incorrect.

Question 18:

Two circles with centers O and P intersect at points A and B. The radii of the circles are 5 cm and 3 cm, respectively, and the distance between their centers is 4 cm. A student states that the length of the common chord AB is 6 cm. Justify or refute this statement with proper reasoning.

Answer:

Let's analyze the given situation step-by-step to verify the student's claim.


Given:

  • Radius of circle with center O (r₁) = 5 cm
  • Radius of circle with center P (r₂) = 3 cm
  • Distance between centers (d) = OP = 4 cm

Step 1: Check if the circles intersect.


Since |r₁ - r₂| < d < r₁ + r₂ (i.e., 2 < 4 < 8), the circles intersect at two points.

Step 2: Calculate the length of the common chord AB.


Using the formula for the length of the common chord:
AB = 2√(r₁² - ( (d² + r₁² - r₂²) / 2d )² )

Step 3: Substitute the values.


AB = 2√(5² - ( (4² + 5² - 3²) / (2 × 4) )² )
= 2√(25 - ( (16 + 25 - 9) / 8 )² )
= 2√(25 - (32/8)² )
= 2√(25 - 16)
= 2√9 = 6 cm

Conclusion: The student's claim is correct. The length of the common chord AB is indeed 6 cm.

Question 19:

In a park, a circular fountain has a radius of 7 meters. A gardener wants to plant flowers around the fountain such that the flowers are always 1 meter away from the edge of the fountain. Calculate the area where the flowers will be planted. Use π = 22/7.

Answer:

Given: Radius of fountain (r) = 7 m, Distance of flowers from edge = 1 m.


Step 1: Calculate the radius of the outer circle (fountain + planting area).
Outer radius (R) = Fountain radius + Distance = 7 m + 1 m = 8 m.


Step 2: Find the area of the outer circle.
Area = πR² = (22/7) × 8 × 8 = 1408/7 ≈ 201.14 m².


Step 3: Find the area of the fountain.
Area = πr² = (22/7) × 7 × 7 = 154 m².


Step 4: Calculate the planting area by subtracting the fountain area from the outer circle area.
Planting area = 201.14 m² - 154 m² = 47.14 m².


Final Answer: The area where the flowers will be planted is 47.14 m².

Question 20:

Two circles with centers O and P have radii 5 cm and 3 cm respectively. The distance between their centers is 10 cm. Determine the length of the direct common tangent between the two circles.

Answer:

Given: Radius of first circle (r₁) = 5 cm, Radius of second circle (r₂) = 3 cm, Distance between centers (d) = 10 cm.


Step 1: Recall the formula for the length of the direct common tangent (L) between two circles:
L = √(d² - (r₁ + r₂)²).


Step 2: Substitute the given values into the formula.
L = √(10² - (5 + 3)²) = √(100 - 64) = √36 = 6 cm.


Final Answer: The length of the direct common tangent is 6 cm.


Note: A direct common tangent touches both circles externally and does not cross the line joining their centers.

Question 21:
A circular park of radius 20 m has a pathway of uniform width around it. The area of the pathway is 264 m². Using this information, answer the following:
(i) Find the width of the pathway.
(ii) Calculate the area of the park including the pathway.
Answer:

Let’s solve the problem step-by-step:


(i) Width of the pathway:
Let the width of the pathway be x meters.
Radius of the park (inner circle) = 20 m.
Radius of the park including pathway (outer circle) = (20 + x) m.
Area of the pathway = Area of outer circle - Area of inner circle.
264 = π(20 + x)² - π(20)².
264 = π[(20 + x)² - 400].
264 = π[400 + 40x + x² - 400].
264 = π(40x + x²).
Taking π ≈ 22/7, we get:
264 = (22/7)(40x + x²).
Multiply both sides by 7: 1848 = 22(40x + x²).
Divide by 22: 84 = 40x + x².
Rearrange: x² + 40x - 84 = 0.
Solve the quadratic equation using factorization:
(x + 42)(x - 2) = 0.
Thus, x = 2 m (since width cannot be negative).

(ii) Area of the park including pathway:
Outer radius = 20 + 2 = 22 m.
Area = π(22)² = (22/7) × 484 ≈ 1521.14 m².

Thus, the width of the pathway is 2 meters, and the total area including the pathway is approximately 1521.14 m².

Question 22:
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:

Here’s the step-by-step solution:


Given:
Radius of smaller circle (r) = 5 cm.
Radius of larger circle (R) = 13 cm.

Step 1: Draw the figure as per the problem.
The chord of the larger circle touches the smaller circle, meaning it is tangent to the smaller circle.

Step 2: The tangent is perpendicular to the radius at the point of contact.
Thus, the radius of the smaller circle to the point of contact forms a right angle with the chord.

Step 3: Let the chord be AB and the point of contact be P.
In the right triangle OAP (where O is the center):
OP = 5 cm (radius of smaller circle).
OA = 13 cm (radius of larger circle).

Step 4: Apply the Pythagorean theorem:
AP² + OP² = OA².
AP² + 25 = 169.
AP² = 144.
AP = 12 cm.

Step 5: Since P is the midpoint of AB (as the perpendicular from the center bisects the chord),
Length of chord AB = 2 × AP = 2 × 12 = 24 cm.

Thus, the length of the chord is 24 cm.

Question 23:
A circular park of radius 20 m has a path of uniform width 5 m around it. Find the area of the path. Use π = 3.14.
Answer:

To find the area of the path around the circular park, follow these steps:


Step 1: Calculate the area of the inner circle (park).
Radius of inner circle (r) = 20 m
Area = πr² = 3.14 × (20)² = 3.14 × 400 = 1256 m²

Step 2: Calculate the radius of the outer circle (park + path).
Width of path = 5 m
Radius of outer circle (R) = 20 m + 5 m = 25 m

Step 3: Calculate the area of the outer circle.
Area = πR² = 3.14 × (25)² = 3.14 × 625 = 1962.5 m²

Step 4: Subtract the area of the inner circle from the outer circle to find the area of the path.
Area of path = 1962.5 m² - 1256 m² = 706.5 m²

Thus, the area of the path is 706.5 m².

Question 24:
Two circles with radii 6 cm and 9 cm have their centers 15 cm apart. Prove that these circles intersect at two distinct points.
Answer:

To prove that the two circles intersect at two distinct points, we use the following conditions based on the distance between centers (d) and their radii (r₁ and r₂):


Given:
Radius of first circle (r₁) = 6 cm
Radius of second circle (r₂) = 9 cm
Distance between centers (d) = 15 cm

Condition for intersection:
Two circles intersect at two distinct points if the distance between their centers is less than the sum of their radii and greater than the difference of their radii.
Mathematically: |r₁ - r₂| < d < r₁ + r₂

Step 1: Calculate the sum of radii.
r₁ + r₂ = 6 cm + 9 cm = 15 cm

Step 2: Calculate the absolute difference of radii.
|r₁ - r₂| = |6 cm - 9 cm| = 3 cm

Step 3: Compare d with the sum and difference.
Here, d = 15 cm, which is equal to the sum of radii (15 cm) but greater than the difference (3 cm).

Since d = r₁ + r₂, the circles touch each other externally at exactly one point. However, if d were slightly less than 15 cm, they would intersect at two points. The given condition suggests the circles are just touching, but the problem states they intersect at two points, implying a slight error in the given data. For two distinct intersection points, d must satisfy 3 cm < d < 15 cm.

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