Circles, Circumcircles and Circumcentre – Step-by-Step NCERT Solutions
Prepare with complete
Class 9 Maths Chapter 5 Exercise 5.1 Solutions
based on the latest
NCERT Ganita Manjari (2026).
This exercise focuses on
constructing triangles and their circumcircles, locating the circumcentre,
understanding whether the circumcentre lies
inside or outside a triangle,
and finding the
least possible radius of a circle passing through two given points.
Every solution is presented in a simple
step-by-step CBSE answer-writing style
to make construction-based circle problems easier to understand and practise.
📖
Exercise
5.1
❓
Questions
4 Questions
📚
Coverage
Circles & Circumcircles
🧠
Skills
Construction & Reasoning
⏱️
Study Time
45–60 Min
⭐
Difficulty
Moderate
🎯
Exam Importance
★★★★★
🎯 By the End of This Exercise, You Will Be Able To…
✅ Construct a Circumcircle of a Triangle
✅ Locate the Circumcentre Using Perpendicular Bisectors
✅ Identify the Position of the Circumcentre
✅ Understand Circumcircle and Equal Radii
✅ Find the Least Possible Radius Through Two Points
Class 9 Maths Chapter 5 Exercise 5.1 Solutions cover the
first exercise of Chapter 5 – I’m Up and Down, and Round and Round
from the latest NCERT Ganita Manjari (2026). This exercise
begins the chapter’s study of circles through geometric constructions and
observations involving triangles, circumcircles, and the circumcentre.
The questions help students develop a clear understanding of these ideas
through practical geometric work.
This page provides clear, question-wise explanations based on the
NCERT Ganita Manjari (2026) textbook and follows a
student-friendly CBSE approach. The solutions are arranged
systematically to support classroom learning, homework, self-study, revision,
and examination preparation while keeping the focus on the actual requirements
of Exercise Set 5.1.
🎯 Exercise Snapshot
📘 What You’ll Find
Complete Exercise Set 5.1 coverage.
NCERT-based question-wise solutions.
Clear explanations of the constructions and observations required.
📚 Page Includes
Question-wise step-by-step solutions.
Triangle and circumcircle construction guidance.
Simple CBSE answer-writing approach.
🏆 Best For
Understanding the beginning of Chapter 5.
Homework and classroom practice.
Revision and CBSE examination preparation.
🎯 Jump to Any Question
Quickly jump to any question of
Class 9 Maths chapter 5 Exercise 5.1 Solutions.
Each card highlights the main concept or task covered in the question,
so you can quickly reach the solution you need.
Understand the basic concepts of circles before solving Exercise Set 5.1.
What is a Circle?
A circle is the set of all points in a plane that are
at the same distance from a fixed point.
The fixed point is called the
centre of the circle.
The common distance is called the
radius.
OA = OB = OC = OD
Centre of a Circle
The fixed point from which every point on the circle is at the
same distance is called the centre.
Remember:
The centre is a point, not a line segment.
Radius
A radius is the line segment joining the
centre of a circle to any point on the circle.
OA is a radius.
A circle has many radii, but all radii of the same circle have
equal length.
Chord
A chord is a line segment whose two endpoints
lie on the circle.
Key point:
A chord does not have to pass through the centre.
Diameter
A diameter is a chord that passes through
the centre of the circle.
Diameter = chord passing through the centre
Chord vs Diameter
Chord
A line segment joining any two points on the circle.
Centre condition:
It may or may not pass through the centre.
Diameter
A chord whose two endpoints lie on the circle and which
passes through the centre.
Centre condition:
It always passes through the centre.
Every diameter is a chord, but every chord is not a diameter.
Relation Between Diameter and Radius
If AB is a diameter and O is the centre, then O divides the
diameter into two radii.
Since AB is a diameter,
AB = AO + OB
Since AO and OB are radii of the same circle,
AO = OB = r
Therefore,
AB = r + r
Hence,
Diameter = 2 × Radius
d = 2r
Therefore,
r = d/2
⭐ The One Idea to Remember
All the basic terms of a circle are connected to one central idea:
Every point on a circle is at the same distance from its centre.
Centre → Radius → Chord → Diameter
⚠️ Common Mistakes
Calling every chord a diameter.
Forgetting that a diameter must pass through the centre.
Thinking a circle has only one radius.
Confusing the radius with the diameter.
Writing d = r instead of
d = 2r.
Quick Check
Before moving ahead, make sure you can answer these questions:
✓ What is a circle?
✓ What is the centre?
✓ What is a radius?
✓ What is a chord?
✓ What makes a chord a diameter?
✓ What is the relation between diameter and radius?
📐 Perpendicular Bisector — The Key Tool
The simple idea that helps us find the centre of a circle.
Perpendicular Bisector
A perpendicular bisector is a line that
cuts a line segment into two equal parts
and is also perpendicular to it.
① It bisects AB
The midpoint E divides AB into two equal parts.
AE = EB
② It is perpendicular
The angle made with AB at E is a right angle.
PE ⟂ AB
AE = EB
and
PE ⟂ AB
Why Are the Distances Equal?
This is the most important property to remember.
Every point on the perpendicular bisector
is equally far from A and B.
For example, if P lies on the perpendicular bisector:
PA = PB
P is on the perpendicular bisector
↓
PA = PB
Theorem: Equal Distances
Statement
Every point on the perpendicular bisector of a line segment
is equidistant from its two endpoints.
Given:
PE is the perpendicular bisector of AB.
To prove:
PA = PB
Proof
Since PE is the perpendicular bisector of AB,
AE = EB
Also, PE is perpendicular to AB.
∠PEA = ∠PEB = 90°
PE is common to both triangles.
PE = PE
Therefore,
△PEA ≅ △PEB
by SAS congruence.
Therefore,
PA = PB
Hence proved.
How Does It Help Find the Centre?
Now comes the important connection with circles.
A and B are on the circle
So OA and OB are both radii.
Radii are equal
OA = OB
Therefore
O is equidistant from A and B.
Therefore, the centre O lies on the
perpendicular bisector of AB.
Constructing a Perpendicular Bisector
Use a ruler and compass.
Follow the four steps carefully.
Step 1 — Draw AB
Draw the given line segment AB.
Step 2 — Arcs from A
With A as centre, draw arcs above and below AB.
Keep the compass opening fixed.
Step 3 — Arcs from B
Without changing the compass opening,
draw arcs from B.
They meet at P and Q.
Step 4 — Join P and Q
Join P and Q.
This line is the perpendicular bisector of AB.
Same compass opening
→
arcs meet at P and Q
→
join P and Q
→
perpendicular bisector
Why Does the Construction Work?
Look at the two points where the equal-radius arcs meet:
P and Q.
At P
PA = PB
At Q
QA = QB
Both P and Q are equally distant from A and B.
Therefore both points lie on the perpendicular bisector.
⭐ Remember This
Circle passes through A and B
↓
OA = OB
↓
O is equidistant from A and B
↓
O lies on the perpendicular bisector of AB
This is the idea you will use in Exercise 5.1.
Whenever two points of a circle are known,
think about their perpendicular bisector.
📚 Circles Through Two Points
Discover why many circles can pass through the same two points.
Can More Than One Circle Pass Through A and B?
Yes. In fact, infinitely many circles can pass
through the same two fixed points A and
B.
The important question is:
Where should their centres be?
We already know the answer:
The centre must lie on the perpendicular bisector of AB.
Where Can the Centre Be?
Let O be the centre of a circle passing through
A and B.
A and B are on the circle
Therefore OA and OB are radii.
All radii are equal
OA = OB
Therefore O is equidistant
O is equally far from A and B.
Therefore, O lies on the perpendicular bisector of AB.
Why Are There Infinitely Many Circles?
The points A and B remain fixed.
But the centre can move to any point on the perpendicular
bisector.
Every new position of the centre gives a new circle through
A and B.
O₁
gives one circle through A and B.
O₂
gives another circle through A and B.
O₃
gives yet another circle through A and B.
The centre can keep moving along the perpendicular bisector.
Therefore, infinitely many circles can pass through A and B.
What Happens When the Centre Moves?
A circle’s radius is the distance from its centre to any
point on the circle.
Centre closer to AB
The radius is smaller.
Centre farther from AB
The radius becomes larger.
Important:
Moving the centre along the perpendicular bisector does not
move A or B. It changes only the radius of the circle.
⭐ Minimum Possible Radius
This idea is especially important for questions such as
Q4 of Exercise 5.1.
Suppose the distance between A and B is fixed.
The centre must lie on the perpendicular bisector.
Why is this the minimum?
The shortest distance from the perpendicular bisector to
either fixed point is obtained when the centre is exactly
at the midpoint E of AB.
Since E is the midpoint of AB,
AE = EB = AB ÷ 2
Therefore,
Minimum radius = AB ÷ 2
🎯 Quick Thinking for Q4
If the distance AB is known:
Step 1
Find the midpoint of AB.
Step 2
Take the midpoint as centre.
Step 3
Radius = half of AB.
Memory rule:
If two fixed points A and B must lie on a circle,
the minimum radius occurs when
the centre is at the midpoint of AB.
⭐ Remember the Whole Idea
Two fixed points A and B
↓
Centre lies on perpendicular bisector
↓
Move the centre
↓
Radius changes
↓
Midpoint of AB → Minimum radius = AB ÷ 2
📚 Circle Through Three Non-Collinear Points
How three points determine one unique circle.
Three Non-Collinear Points
Suppose three points A, B and C are given,
and they are not on the same straight line.
Non-collinear means that A, B and C do not lie
on one straight line.
A, B and C form a triangle.
Why Exactly One Circle?
From the previous concept, we know that the centre of a circle
passing through two points lies on the perpendicular bisector
of the segment joining those points.
For A and B
The centre must lie on the perpendicular bisector of AB.
For A and C
The centre must lie on the perpendicular bisector of AC.
The two bisectors meet once
Their unique intersection is the centre of the circle.
Key idea:
Since the triangle is non-collinear, the perpendicular bisectors
of two sides are not parallel. They meet at exactly one point.
That point is the circumcentre.
Circumcentre and Circumcircle
📍 Circumcentre
The common point where the perpendicular bisectors
of the sides of a triangle meet is called its
circumcentre.
⭕ Circumcircle
The circle passing through all three vertices
A, B and C is called the circumcircle
of the triangle.
The perpendicular bisectors of AB and AC meet at O.
OA = OB = OC
OA = OB = OC
Therefore one circle centred at O passes through A, B and C.
⭐ Theorem
Statement
The perpendicular bisectors of the sides of a triangle
are concurrent.
Their common point is equidistant from the
three vertices of the triangle.
OA = OB = OC
Step-by-Step Proof
Given:
ABC is a non-collinear triangle.
The perpendicular bisectors of AB and AC meet at O.
To prove:
OA = OB = OC
1
O lies on the perpendicular bisector of AB.
OA = OB
2
O also lies on the perpendicular bisector of AC.
OA = OC
3
From the two equalities,
OA = OB
and
OA = OC
Therefore,
OA = OB = OC
4
Thus O is equally distant from A, B and C.
Therefore a circle with centre O and radius OA
passes through all three points.
Hence the circle through A, B and C exists.
And it is unique.
Construction: Circumcentre and Circumcircle
Given a triangle ABC, we can construct its circumcircle
using the perpendicular bisectors of two sides.
Step 1 — Draw triangle ABC
Draw the given non-collinear triangle ABC.
Step 2 — Bisect AB
Construct the perpendicular bisector of AB using
equal-radius arcs.
Step 3 — Bisect AC
Construct the perpendicular bisector of AC.
Let the two bisectors meet at O.
Step 4 — Draw the circle
Take O as centre and OA as radius.
Draw the circle.
Draw triangle
→ Bisect AB
→ Bisect AC
→ Their intersection is O
→ Take OA as radius
→ Draw circumcircle
Why Are Two Perpendicular Bisectors Enough?
We do not need to construct all three perpendicular bisectors.
If O lies on the perpendicular bisector of AB,
OA = OB
If O lies on the perpendicular bisector of AC,
OA = OC
Therefore,
OA = OB = OC
So the circle centred at O automatically passes through
all three vertices.
⭐ Remember This for Exercise 5.1
Three non-collinear points
↓
Perpendicular bisectors of two sides
↓
Common point O
↓
OA = OB = OC
↓
One unique circumcircle
Exam connection:
The point where the perpendicular bisectors meet is the
circumcentre, and the circle through all
three vertices is the circumcircle.
📍 Position of the Circumcentre
The position of the circumcentre depends on the type of triangle.
Acute Triangle
All three angles are less than 90°.
Circumcentre O → INSIDE
Right Triangle
One angle is exactly 90°.
O → Midpoint of the Hypotenuse
Obtuse Triangle
One angle is greater than 90°.
Circumcentre O → OUTSIDE
Acute → Inside
|
Right → Midpoint of Hypotenuse
|
Obtuse → Outside
📐 How to Construct a Triangle with Given Measurements
Learn the correct construction method for SSS, SAS and ASA cases.
SSS
Three sides given
SAS
Two sides + included angle
ASA
Two angles + included side
QUESTION 1 — SSS
Construct a triangle ABC in which:
AB = 6 cm |
BC = 5 cm |
CA = 4 cm
Solution — Construction
Given:
AB = 6 cm, BC = 5 cm, CA = 4 cm
Method: SSS
All three sides are given, so we use the
SSS construction.
Step 1 — Draw BC
Draw BC = 5 cm.
Step 2 — Arc from B
With B as centre and radius
6 cm, draw an arc.
Step 3 — Draw the Arc from C
Now change the compass opening to
4 cm. Take C as centre
and draw an arc. It intersects the first arc at
A.
Step 4 — Join A to B and C
Join AB and AC.
✓ Required Triangle Constructed
Therefore, △ABC is the required triangle.
SSS:
Draw one side → draw an arc from each endpoint
→ their intersection gives the third vertex
→ join the sides.
QUESTION 2 — SAS
Construct a triangle ABC in which:
AB = 6 cm |
AC = 5 cm |
∠BAC = 60°
Solution — Construction
Given:
AB = 6 cm, AC = 5 cm, ∠BAC = 60°
Method: SAS
The two given sides meet at A, and the angle between them
is given. Therefore, we use the SAS construction.
Step 1 — Draw AB
Draw AB = 6 cm.
Step 2 — Construct 60° at A
At A, construct ∠BAC = 60°
and draw the required ray.
Step 3 — Mark C
On the ray, mark AC = 5 cm.
Mark this point as C.
Step 4 — Join BC
Join BC to complete the triangle.
✓ Required Triangle Constructed
Therefore, △ABC is the required triangle.
SAS:
Draw one side → construct the included angle
→ mark the second side → join the remaining vertices.
QUESTION 3 — ASA
Construct a triangle ABC in which:
BC = 6 cm |
∠ABC = 50° |
∠ACB = 60°
Solution — Construction
Given:
BC = 6 cm, ∠ABC = 50°, ∠ACB = 60°
Method: ASA
The side BC and the two angles at its
endpoints are given. Therefore, we use the
ASA construction.
Step 1 — Draw BC
Draw BC = 6 cm.
Step 2 — Construct 50° at B
At B, construct ∠ABC = 50°
and draw the ray.
Step 3 — Construct 60° at C
At C, construct ∠ACB = 60°.
The two rays meet at A.
Step 4 — Complete the Triangle
The two rays meet at A.
Thus, △ABC is constructed.
✓ Required Triangle Constructed
Therefore, △ABC is the required triangle.
ASA:
Draw the given side → construct the given angle
at each endpoint → the two rays meet at the third vertex.
🎯 Let’s Put It All Together!
One complete model problem — from constructing the triangle
to finding its circumcentre and circumcircle.
👨🏫 Rakesh Sir says:
“Ab hum ek complete question karenge.
Pehle triangle banayenge, phir uska circumcentre
find karenge, aur finally circumcircle draw karenge.
Agar ye samajh aa gaya, Exercise 5.1 kaafi easy lagegi!” 😊
MODEL QUESTION
Construct a triangle ABC such that:
AB = 5 cm
|
BC = 6 cm
|
CA = 7 cm
Then construct its circumcentre and
circumcircle. Also state the position
of the circumcentre.
🤔 Before We Start — Think!
Look carefully at the information given in the question.
AB = 5 cm
BC = 6 cm
CA = 7 cm
Three sides are given.
💡 Therefore → SSS Construction
SSS = Side – Side – Side
🧭 Our Plan
Draw Triangle
→
Find Circumcentre
→
Draw Circumcircle
🚀 Ready? Let’s construct the triangle first!
In the next step, we will use the three given sides
to construct △ABC.
✏️ Step 1 — Construct the Triangle
Let’s use the SSS method to construct △ABC.
Given:
AB = 5 cm |
BC = 6 cm |
CA = 7 cm
Step 1 — Draw BC
Draw a line segment
BC = 6 cm.
Step 2 — Arc from B
With B as centre and radius
5 cm, draw an arc.
Step 3 — Arc from C
With C as centre and radius
7 cm, draw another arc.
Step 4 — Join A to B and C
The arcs meet at A.
Join AB and AC.
🎉 Triangle ABC is ready!
We used the three given sides to construct the triangle.
Next, we will find its circumcentre. 🎯
🎯 Step 2 — Find the Circumcentre
The perpendicular bisectors will help us locate the centre.
👨🏫 Remember the trick!
The centre of a circle passing through two points
must lie on the perpendicular bisector
of the segment joining those points.
Step 1 — Bisect AB
Construct the perpendicular bisector of AB.
Every point on this line is equally distant
from A and B.
Step 2 — Bisect BC
Now construct the perpendicular bisector of BC.
Every point on this line is equally distant
from B and C.
Step 3 — Meet at O
The two perpendicular bisectors meet at
O.
🎯 O is the Circumcentre
💡 Why does O become the centre?
O lies on the perpendicular bisector of AB,
so:
OA = OB
O also lies on the perpendicular bisector of BC,
so:
OB = OC
Therefore, OA = OB = OC
🎉 We found the centre!
Now that we know O, the next step is easy:
we will use O as the centre to draw the
circumcircle. ⭕
⭕ Step 3 — Draw the Circumcircle
We know the centre. Now let’s use it to draw the circle.
🤔 What do we need now?
We already know the centre O.
Which point can we use to get the radius?
💡 OA
Step 1 — Take OA as the Radius
Place the compass point at O and open it
up to A.
Radius = OA
Step 2 — Draw the Circle
Keeping O fixed as the centre,
draw the complete circle.
🎯 Look What Happened!
The same radius OA reaches all three
vertices of the triangle.
OA = OB = OC
So A, B and C all lie on the same circle.
👨🏫 Rakesh Sir says:
“Centre mil gaya, radius mil gaya —
बस circle draw kar do!” 😊
This circle is called the
circumcircle of △ABC.
Next:
Let’s check where the circumcentre lies
and verify our construction. 🎯
✅ Position & Verification
Let’s check our construction and understand where O lies.
🤔 One last thing!
We have found the circumcentre O.
Now, where is it located?
The sides of the triangle are
5 cm, 6 cm and 7 cm.
This forms an acute triangle.
🎯 Circumcentre O lies INSIDE the triangle.
🔎 Our Complete Construction
O is clearly inside the acute triangle.
🔍 Let’s Verify
✓ OA = OB
✓ OB = OC
✓ OA = OC
✓ Circle passes through A
✓ Circle passes through B
✓ Circle passes through C
🧠 Remember the Complete Process
3 sides given
→
Construct △ABC
→
Perpendicular bisectors
→
O = Circumcentre
→
Draw circumcircle
🎉 You Got It!
👨🏫 “Triangle banana → perpendicular bisectors →
O mil gaya → radius OA liya →
circle draw kar diya!” 😊
Now you are ready to tackle the construction questions
in Exercise 5.1. 💪
📝 Class 9 Maths Chapter 5 Exercise 5.1 Solutions
Solve every question of Class 9 Maths Chapter 5 Exercise 5.1
with clear, step-by-step NCERT Ganita Manjari (2026) solutions.
This exercise introduces important geometric ideas involving triangles, circumcircles, perpendicular bisectors, circumcentres,
and related constructions. Each solution is presented in a simple,
student-friendly CBSE answer-writing style with the required
construction steps and reasoning, making it useful for classroom work,
homework, revision, and examination preparation.
Draw △ABC with AB = 5 cm, ∠A = 70° and ∠B = 60°.
Draw the circumcircle of △ABC. Is the centre inside or outside
the triangle?
✍️ Solution
Given,
AB = 5 cm
∠A = 70°
∠B = 60°
To construct,
1. △ABC
2. The circumcircle of △ABC
3. The position of its circumcentre
Construction:
First construct △ABC using the given side and angles.
Then construct the perpendicular bisectors to locate the
circumcentre and draw the circumcircle.
📐 Construction Figure
Figure: Construction of △ABC, its circumcircle
and perpendicular bisectors meeting at O.
✏️ Construction
Step 1.
Draw AB = 5 cm.
Step 2.
At A, construct ∠BAC = 70° and draw ray AC.
Step 3.
At B, construct ∠ABC = 60° and draw ray BC.
Step 4.
Let the rays AC and BC meet at C. Thus, △ABC is constructed.
Step 5.
Construct the perpendicular bisector of AB.
Step 6.
Construct the perpendicular bisector of AC.
Step 7.
Let the two perpendicular bisectors meet at O.
Then O is the circumcentre of △ABC.
Step 8.
With O as centre and OA as radius,
draw the circumcircle.
🔍 Observation
∠C = 180° − 70° − 60° = 50°.
All three angles of △ABC are less than 90°. Therefore, △ABC
is an acute-angled triangle, and its circumcentre lies
inside the triangle.
✅ Final Answer
The circumcircle of △ABC is constructed successfully.
Its centre O lies inside the triangle.
Question 2
Draw △ABC with AB = 5 cm, ∠A = 100°, AC = 4 cm.
Draw the circumcircle of △ABC. Is the centre inside or outside
the triangle?
✍️ Solution
Given,
AB = 5 cm
∠A = 100°
AC = 4 cm
To construct,
1. △ABC
2. The circumcircle of △ABC
3. The position of its circumcentre
Construction:
First construct △ABC using the given side, angle and side.
Then construct the perpendicular bisectors to locate the
circumcentre and draw the circumcircle.
📐 Construction Figure
Figure: Triangle ABC with its circumcircle,
perpendicular bisectors and circumcentre O.
✏️ Construction
Step 1.
Draw AB = 5 cm.
Step 2.
At A, construct ∠BAC = 100° and draw ray AC.
Step 3.
On ray AC, mark C such that AC = 4 cm.
Step 4.
Join BC. Thus, △ABC is constructed.
Step 5.
Construct the perpendicular bisector of AB.
Step 6.
Construct the perpendicular bisector of AC.
Step 7.
Let the two perpendicular bisectors meet at O.
Then O is the circumcentre of △ABC.
Step 8.
With O as centre and OA as radius,
draw the circumcircle.
🔍 Observation
Given,
∠A = 100°, which is greater than 90°.
Therefore, △ABC is an obtuse-angled triangle.
The circumcentre of an obtuse-angled triangle lies
outside the triangle.
✅ Final Answer
The circumcircle of △ABC is constructed successfully.
Its centre O lies outside the triangle.
Question 3
Draw △ABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm.
Draw the circumcircle of △ABC. Let the circumcentre be O.
Measure OA, OB, OC.
✍️ Solution
Given,
AB = 6 cm
BC = 7 cm
CA = 7 cm
To construct,
1. △ABC
2. The circumcircle of △ABC
3. Measure OA, OB and OC
Construction:
First construct △ABC using the three given sides.
Then construct two perpendicular bisectors to locate the
circumcentre O and draw the circumcircle.
📐 Construction Figure
Figure: Triangle ABC with its circumcircle
and circumcentre O.
✏️ Construction
Step 1.
Draw AB = 6 cm.
Step 2.
With A as centre and radius 7 cm, draw an arc.
Step 3.
With B as centre and radius 7 cm, draw another arc
intersecting the first arc at C.
Step 4.
Join AC and BC. Thus, △ABC is constructed.
Step 5.
Construct the perpendicular bisector of AB.
Step 6.
Construct the perpendicular bisector of AC.
Step 7.
Let the two perpendicular bisectors meet at O.
Then O is the circumcentre of △ABC.
Step 8.
With O as centre and OA as radius,
draw the circumcircle.
📏 Measurement
On measuring the three radii of the circumcircle,
OA ≈ 3.9 cm OB ≈ 3.9 cm OC ≈ 3.9 cm
🔍 Observation
The measured values of OA, OB and OC are equal.
Hence, O is equidistant from A, B and C and is therefore the
circumcentre of △ABC.
✅ Final Answer
OA ≈ OB ≈ OC ≈ 3.9 cm.
Question 4
What is the least possible radius of a circle through two
points A and B?
✍️ Solution
Given,
Two fixed points A and B.
📐 Geometrical Figure
Figure:
Different circles through A and B have their centres
M, O₁ and O₂ on the perpendicular
bisector of AB.
🔍 Explanation
Step 1.
Let M be the midpoint of AB.
AM = MB = AB/2
Step 2.
The centre of every circle passing through A and B lies on the
perpendicular bisector of AB.
Step 3.
If M is the centre, then
r = MA = MB = AB/2
Thus, AB is the diameter.
Step 4.
For another centre O₁, triangle AMO₁ is right-angled at M.
Therefore, r₁ = O₁A is the hypotenuse and
r₁ > AM = r
Step 5.
Similarly, for centre O₂, triangle AMO₂ is right-angled at M.
Its hypotenuse is r₂ = O₂A. Hence,
r₂ > AM = r
Also, since O₂ is farther from AB than O₁,
r₂ > r₁.
Step 6.
Therefore,
r < r₁ < r₂
So the smallest radius is r = AB/2, obtained when
the centre is at the midpoint M of AB.
📌 Conclusion
The smallest possible radius is obtained when the centre is the
midpoint of AB. In this case, AB is the diameter.
✅ Final Answer
Least possible radius = AB/2
📚 Continue Learning
Congratulations! You have completed the
Class 9 Maths chapter 5 Exercise 5.1 Solutions.
You have now worked through the first exercise of
Chapter 5 – I’m Up and Down, and Round and Round.
Continue your learning journey by moving to the next exercise,
exploring the complete chapter, or revisiting Exercise 5.1 whenever
you need a quick revision.
Revise the complete Class 9 Maths Chapter 5 — Circles
in a few minutes before solving Exercise 5.1.
Think of this as your whole lesson in short. 🎯
⭕ Circle — Know the Basics
✔ Centre: fixed point from which the
circle is drawn.
✔ Radius: distance from the centre
to any point on the circle.
✔ Chord: joins any two points on
the circle.
✔ Diameter: chord passing through
the centre.
💡 Remember:
Diameter = 2 × Radius
📏 Perpendicular Bisector
✔ Divides a line segment into
two equal parts at 90°.
✔ Every point on it is
equidistant from the endpoints.
✔ For segment AB, a point O on its perpendicular
bisector gives:
OA = OB.
🎯 Key use:
It helps us locate the centre of a circle.
🎯 Circumcentre — Find It
✔ In △ABC, construct the perpendicular bisectors
of any two sides.
✔ Their intersection is the
circumcentre O.
✔ Then:
OA = OB = OC.
💡 Don’t guess the centre — construct it.
⭕ Circumcircle — Draw It
✔ Take O as the centre.
✔ Take OA (or OB or OC) as the radius.
✔ Draw the circle through
A, B and C.
🎯 This circle is called the
circumcircle.
📍 Position of Circumcentre
✔ Acute triangle
→ Circumcentre is inside.
✔ Right triangle
→ Circumcentre is the
midpoint of hypotenuse.
✔ Obtuse triangle
→ Circumcentre is outside.
🛠️ Triangle Construction Skills
✔ SSS:
Three sides given → use two arcs.
✔ SAS:
Two sides + included angle → construct the angle,
then mark the second side.
✔ ASA:
Two angles + included side → construct both endpoint
angles; their rays meet at the third vertex.
💡 Always follow the measurements carefully.
🧠 If You Forget Everything, Remember This
Construct triangle
→
Two perpendicular bisectors
→
Intersection = O
→
O = Circumcentre
→
OA as radius
→
Draw circumcircle
✅ Final Exercise 5.1 Checklist
☑ I know circle, radius, chord and diameter.
☑ I can construct a perpendicular bisector.
☑ I can locate the circumcentre.
☑ I know the position of O in all three triangle types.
☑ I can construct a circumcircle.
☑ I can construct triangles using SSS, SAS and ASA.
☑ I know why OA = OB = OC.
🚀 I’m ready for Exercise 5.1!
❓ Frequently Asked Questions
Find answers to common questions about
Class 9 Maths Exercise 5.1 Solutions.
These FAQs are designed to help you understand the ideas behind the exercise,
avoid common construction mistakes, and prepare confidently for
Chapter 5 – I’m Up and Down, and Round and Round.
What is the main focus of Exercise Set 5.1?
Exercise Set 5.1 introduces the construction-based ideas of
Chapter 5 – I’m Up and Down, and Round and Round.
The four questions involve constructing triangles, drawing their circumcircles,
identifying the position of the circumcentre, measuring the radii of a
circumcircle, and understanding the least possible radius of a circle through
two given points.
What should I know before attempting Exercise 5.1?
You should be familiar with basic triangle construction, perpendicular
bisectors, circles, radius, and the idea of a circumcircle and circumcentre.
It is also important to understand that the perpendicular bisectors of the
sides of a triangle meet at its circumcentre. These ideas are developed in
the beginning of Chapter 5 before Exercise Set 5.1.
How do I construct the circumcircle of a triangle?
First construct the given triangle. Then construct the perpendicular
bisectors of any two of its sides. Their point of intersection is the
circumcentre. Taking this point as the centre and the distance
to any vertex as the radius, draw the circle. The circle passes through all
three vertices of the triangle and is called its circumcircle.
Where does the circumcentre lie in different types of triangles?
For an acute-angled triangle, the circumcentre lies inside the
triangle. For an obtuse-angled triangle, it lies outside the
triangle. For a right-angled triangle, the circumcentre lies at
the midpoint of the hypotenuse.
Why are OA, OB and OC equal in Question 3?
In Question 3, O is the circumcentre of triangle ABC. Therefore A, B and C
all lie on the same circumcircle with centre O. Since the radii of the same
circle are equal, the distances from O to A, B and C are equal. Hence,
OA = OB = OC.
What is the least possible radius of a circle passing through two points?
For two given points A and B, infinitely many circles can pass through them.
The least possible radius is obtained when the centre is at the midpoint of
AB. In that case, the radius is the distance from the midpoint of AB to
either A or B.
What are the common mistakes in Exercise Set 5.1?
Common mistakes include constructing only one perpendicular bisector,
incorrectly locating the circumcentre, drawing the circumcircle with an
incorrect radius, and confusing the position of the circumcentre in acute
and obtuse triangles. Students should also take care to use accurate
construction and measurement.
Are these Class 9 Maths Exercise 5.1 Solutions based on Ganita Manjari 2026?
Yes. The solutions on this page are prepared from
Chapter 5 – I’m Up and Down, and Round and Round of the
NCERT Ganita Manjari (2026) Grade 9 textbook. Exercise Set
5.1 contains four questions covering triangle construction, circumcircles,
circumcentre, measurement of radii, and the least possible radius through two
given points.
📚 Useful Learning Resources
Continue your preparation with more
Class 9 Maths resources from Maths Gurukulam,
or visit the official NCERT and CBSE websites for the latest textbooks,
syllabus, and academic updates.
Find Class 9 Maths Chapter 5 Exercise 5.1 Solutions helpful?
For students who need personal support beyond self-study, join
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These Class 9 Maths Chapter 5 Exercise 5.1 Solutions are
carefully prepared according to the latest NCERT Ganita Manjari
(2026) and the CBSE curriculum. The solutions follow
a clear, step-by-step approach to help students understand the construction
and reasoning used in Exercise 5.1 and build confidence in circle geometry.
