A circle is the set of all points in a plane that are at a fixed distance from a fixed point. The fixed point is the centre, and the fixed distance is the radius.
Important questions class 9 maths chapter 5 help students practise circle definitions, chords, perpendicular bisectors, circumcircle, arcs, angles in the same segment, angle in a semicircle and cyclic quadrilaterals. This chapter builds geometric proof skills, so students should revise each theorem with a diagram and reason.
Class 9 Maths Chapter 5 begins with circles seen in nature, such as raindrops on water, the full moon, the Sun and plant stems. It then turns this observation into a mathematical definition: every point on a circle lies at the same distance from its centre.
The chapter teaches students how circles behave through symmetry, chords and arcs. Students learn why three non-collinear points determine one circle, why equal chords subtend equal angles, why a perpendicular from the centre bisects a chord, and why opposite angles of a cyclic quadrilateral add to 180°. These results appear often in proof-based and numerical geometry questions.
Key Takeaways from Class 9 Maths Chapter 5
| Topic |
What Students Must Know |
| Chapter Name |
I’m Up and Down, and Round and Round |
| Main Concept |
Circles and their properties |
| Core Terms |
Centre, radius, chord, diameter, arc |
| Key Construction |
Circumcircle of a triangle |
| Main Theorems |
Chord, arc and cyclic quadrilateral theorems |
| Important Formula |
Chord length = (2\sqrt{r^2-d^2}) |
| Exam Focus |
Proofs, diagrams and numerical problems |
Important Questions Class 9 Maths Chapter 5 with Answers: Circle Basics
Q1. What is a circle?
A circle is the set of all points in a plane that are at a fixed distance from a fixed point.
The fixed point is called the centre. The fixed distance is called the radius.
Q2. What is a chord of a circle?
A chord is a line segment joining any two points on a circle.
A diameter is a special chord that passes through the centre of the circle.
Q3. What is the longest chord of a circle?
The diameter is the longest chord of a circle.
It passes through the centre and has length twice the radius.
Q4. What is an arc of a circle?
An arc is a connected part of a circle between two points on the circle.
The smaller part is called the minor arc. The larger part is called the major arc.
Q5. What is a cyclic quadrilateral?
A cyclic quadrilateral is a quadrilateral whose four vertices lie on the same circle.
In a cyclic quadrilateral, the sum of each pair of opposite angles is 180°.
Class 9 Maths Chapter 5 Question Answer on Circumcircle
The circumcircle concept is important because it connects triangles, perpendicular bisectors and circles. Students should remember that three non-collinear points determine exactly one circle.
Circumcircle Class 9 Questions
Q1. How many circles can pass through two given points?
Infinitely many circles can pass through two given points.
The centres of all such circles lie on the perpendicular bisector of the line segment joining the two points.
Q2. What is the least possible radius of a circle through two points A and B?
The least possible radius is half of (AB).
This happens when (AB) is the diameter of the circle.
Q3. How many circles can pass through three non-collinear points?
Exactly one circle can pass through three non-collinear points.
This circle is called the circumcircle of the triangle formed by those three points.
Q4. Why can no circle pass through three collinear points?
A circle can cut a line at a maximum of two points.
Three collinear points lie on one straight line. So, no circle can pass through all three distinct collinear points.
Q5. What is the circumcentre of a triangle?
The circumcentre is the centre of the circle passing through the three vertices of a triangle.
It lies at the intersection of the perpendicular bisectors of the sides.
Circles Class 9 Important Questions on Chords
Chord theorems are central to this chapter. Students should revise both theorem statements and their converse.
Chords of a Circle Class 9 Theorem Questions
Q1. Show that the triangle formed by a chord and the centre of the circle is isosceles.
Let (AB) be a chord of a circle with centre (O).
Join (OA) and (OB).
Since (OA) and (OB) are radii of the same circle:
OA = OB
Therefore, (△ OAB) is isosceles.
Q2. State the theorem on equal chords and angles at the centre.
Equal chords of a circle subtend equal angles at the centre.
If (AB = CD), then:
∠ AOB = ∠ COD
where (O) is the centre.
Q3. State the converse of the equal chords theorem.
Chords of a circle that subtend equal angles at the centre are equal.
If:
∠ AOB = ∠ COD
then:
AB = CD
Q4. What is the relation between a chord and the perpendicular from the centre?
The perpendicular from the centre of a circle to a chord bisects the chord.
If (OM ⊥ AB), then:
AM = MB
Q5. What happens when a line joins the centre of a circle to the midpoint of a chord?
The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.
If (M) is the midpoint of chord (AB), then:
OM ⊥ AB
Chords of a Circle Class 9 Numerical Questions
These questions test theorem use with the Baudhāyana-Pythagoras theorem. Draw the radius to one end of the chord and the perpendicular from the centre.
Q1. Find the length of a chord if the radius is 7 cm and its perpendicular distance from the centre is 6 cm.
Let radius (r = 7) cm.
Distance from centre (d = 6) cm.
Half chord:
= √(r^2 - d^2)
= √(7^2 - 6^2)
= √(49 - 36)
= √(13)
Chord length:
= 2√(13) { cm}
Q2. In a circle, a chord is 5 cm away from the centre. If the radius is 13 cm, find the chord length.
Here, (r = 13) cm and (d = 5) cm.
Half chord:
= √(13^2 - 5^2)
= √(169 - 25)
= √(144)
= 12
Chord length:
= 2 × 12 = 24 { cm}
Q3. A circle has radius 15 cm. A chord is 9 cm from the centre. Find the chord length.
Here, (r = 15) cm and (d = 9) cm.
Half chord:
= √(15^2 - 9^2)
= √(225 - 81)
= √(144)
= 12
Chord length:
= 24 { cm}
Q4. A chord of length 24 cm is drawn in a circle of diameter 26 cm. Find its distance from the centre.
Diameter = 26 cm.
Radius:
r = 13 { cm}
Chord length = 24 cm.
Half chord:
= 12 { cm}
Distance from centre:
d = √(13^2 - 12^2)
= √(169 - 144)
= √(25)
= 5 { cm}
Q5. The distance of a chord of length 16 cm from the centre is 6 cm. Find the radius.
Half chord:
= 8 { cm}
Distance from centre:
= 6 { cm}
Radius:
r = √(8^2 + 6^2)
= √(64 + 36)
= √(100)
= 10 { cm}
Angle Subtended by Arc Class 9 Important Questions
This section is important for theorem-based geometry questions. Students should remember that the angle at the centre is double the angle at the circle.
Circle Theorems Class 9 on Angles
Q1. State the theorem on angle subtended by an arc.
The angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at any point on the remaining part of the circle.
If arc (AB) subtends (∠ AOB) at the centre and (∠ ACB) at the circle, then:
∠ AOB = 2∠ ACB
Q2. An arc subtends 70° at the centre. Find the angle subtended at a point on the circle.
Angle at centre:
= 70^°
Angle at circle:
= (70^°)/(2)
= 35^°
Q3. If an angle at the circle is 50°, what is the angle subtended by the same arc at the centre?
Angle at centre:
= 2 × 50^°
= 100^°
Q4. What is the angle subtended by a diameter at any point on the circle?
A diameter subtends a right angle at any point on the circle.
So, the angle is:
90^°
Q5. Why is the angle in a semicircle 90°?
A diameter subtends (180^°) at the centre.
The angle at the circle is half of the angle at the centre.
(180^°)/(2) = 90^°
So, the angle in a semicircle is (90^°).
Cyclic Quadrilateral Class 9 Important Questions
Cyclic quadrilaterals are common in exams because they mix theorem recall with angle calculations.
Q1. State the main property of a cyclic quadrilateral.
The sum of opposite angles of a cyclic quadrilateral is (180^°).
If (ABCD) is cyclic, then:
∠ A + ∠ C = 180^°
and
∠ B + ∠ D = 180^°
Q2. If (ABCD) is cyclic and (∠ A = 75^°), find (∠ C).
In a cyclic quadrilateral:
∠ A + ∠ C = 180^°
75^° + ∠ C = 180^°
∠ C = 105^°
Q3. If (ABCD) is cyclic and (∠ B = 110^°), find (∠ D).
∠ B + ∠ D = 180^°
110^° + ∠ D = 180^°
∠ D = 70^°
Q4. Quadrilateral (PQRS) is cyclic. If (∠ P = (2x + 10)^°) and (∠ R = (3x - 20)^°), find (x).
Opposite angles of a cyclic quadrilateral add to (180^°).
(2x + 10) + (3x - 20) = 180
5x - 10 = 180
5x = 190
x = 38
Now:
∠ P = 2(38) + 10 = 86^°
∠ R = 3(38) - 20 = 94^°
Q5. Can a cyclic quadrilateral have angles 80°, 110°, 100° and 70°?
Yes, it can be cyclic if opposite angles add to (180^°).
80^° + 100^° = 180^°
110^° + 70^° = 180^°
So, such a cyclic quadrilateral can be drawn.
Class 9 Maths Chapter 5 MCQ with Answers
Class 9 Maths Chapter 5 MCQ questions test definitions, theorems and quick angle reasoning. Students should revise these after learning the proof steps.
Q1. The longest chord of a circle is:
(a) Radius
(b) Arc
(c) Diameter
(d) Tangent
Answer: (c) Diameter
The diameter passes through the centre and has the greatest chord length.
Q2. A circle has how many lines of reflection symmetry?
(a) One
(b) Two
(c) Four
(d) Infinitely many
Answer: (d) Infinitely many
Every diameter is a line of reflection symmetry.
Q3. How many circles can pass through three non-collinear points?
(a) Zero
(b) One
(c) Two
(d) Infinitely many
Answer: (b) One
Three non-collinear points determine a unique circle.
Q4. Equal chords of a circle subtend:
(a) Unequal angles at the centre
(b) Equal angles at the centre
(c) Right angles only
(d) Straight angles only
Answer: (b) Equal angles at the centre
This is a standard chord theorem.
Q5. In a cyclic quadrilateral, each pair of opposite angles is:
(a) Equal
(b) Complementary
(c) Supplementary
(d) Always 90°
Answer: (c) Supplementary
Opposite angles add to (180^°).
Q6. The angle in a semicircle is always:
(a) (45^°)
(b) (60^°)
(c) (90^°)
(d) (180^°)
Answer: (c) (90^°)
A diameter subtends a right angle at any point on the circle.
Class 9 Maths Chapter 5 Extra Questions with Answers
Class 9 Maths Chapter 5 extra questions help students revise symmetry, circumcentre positions and chord-distance ideas.
Q1. Why is every diameter a line of symmetry of a circle?
A diameter divides the circle into two equal halves.
If the circle is folded along any diameter, both halves overlap exactly. So, every diameter is a line of symmetry.
Q2. Why does a circle have complete rotational symmetry?
A circle looks the same after rotation through any angle about its centre.
Every point remains at the same distance from the centre. So, the circle has complete rotational symmetry.
Q3. Where does the circumcentre lie in an acute-angled triangle?
In an acute-angled triangle, the circumcentre lies inside the triangle.
This happens because the perpendicular bisectors of the sides meet inside the triangle.
Q4. Where does the circumcentre lie in a right-angled triangle?
In a right-angled triangle, the circumcentre lies at the midpoint of the hypotenuse.
The hypotenuse becomes the diameter of the circumcircle.
Q5. Where does the circumcentre lie in an obtuse-angled triangle?
In an obtuse-angled triangle, the circumcentre lies outside the triangle.
The perpendicular bisectors meet outside because one angle is greater than (90^°).
Q6. Which chord is closer to the centre: longer chord or shorter chord?
The longer chord is closer to the centre.
A diameter is the closest chord to the centre because it passes through the centre.
Long Answer Questions from Class 9 Maths Chapter 5
Long answers should include the given statement, construction or figure idea, theorem used and final conclusion.
Proof-Based Circle Theorems Class 9
Q1. Prove that equal chords of a circle subtend equal angles at the centre.
Let (AB) and (CD) be equal chords of a circle with centre (O).
Join (OA), (OB), (OC) and (OD).
Since all are radii of the same circle:
OA = OB = OC = OD
Given:
AB = CD
Now compare (△ AOB) and (△ COD).
OA = OC
OB = OD
AB = CD
So, by SSS congruence:
△ AOB ≅ △ COD
Therefore:
∠ AOB = ∠ COD
Hence, equal chords subtend equal angles at the centre.
Q2. Prove that the perpendicular from the centre to a chord bisects the chord.
Let (AB) be a chord of a circle with centre (O).
Let (OM ⊥ AB).
Join (OA) and (OB).
Since (OA) and (OB) are radii:
OA = OB
In (△ OMA) and (△ OMB):
OA = OB
OM = OM
∠ OMA = ∠ OMB = 90^°
So, by RHS congruence:
△ OMA ≅ △ OMB
Therefore:
AM = MB
Hence, the perpendicular from the centre to a chord bisects the chord.
Q3. Prove that the angle subtended by an arc at the centre is double the angle subtended at the circle.
Let arc (AB) subtend (∠ AOB) at the centre (O) and (∠ ACB) at a point (C) on the remaining part of the circle.
Join (OA), (OB) and (OC).
Since (OA = OC), (△ AOC) is isosceles.
Since (OB = OC), (△ BOC) is also isosceles.
Using the exterior angle theorem in the two isosceles triangles, the central angle becomes twice the angle at the circle.
Therefore:
∠ AOB = 2∠ ACB
Hence proved.
Q4. Prove that the angle in a semicircle is 90°.
Let (AB) be the diameter of a circle with centre (O).
Let (C) be any point on the circle.
The diameter (AB) subtends a straight angle at the centre.
∠ AOB = 180^°
By the arc angle theorem:
∠ ACB = (1)/(2)∠ AOB
∠ ACB = (1)/(2) × 180^°
∠ ACB = 90^°
So, the angle in a semicircle is (90^°).
Q5. Prove that opposite angles of a cyclic quadrilateral are supplementary.
Let (ABCD) be a cyclic quadrilateral.
Since all four points lie on a circle, (∠ A) and (∠ C) stand on opposite arcs.
The central angles for the two arcs together make a full angle of (360^°).
Angles at the circle are half the corresponding central angles.
So:
∠ A + ∠ C = (1)/(2) × 360^°
= 180^°
Similarly:
∠ B + ∠ D = 180^°
Hence, opposite angles of a cyclic quadrilateral are supplementary.
Application-Based Questions from I’m Up and Down, and Round and Round Class 9
These questions test theorem selection. Students should identify whether the problem uses chords, angles or cyclic quadrilaterals.
Q1. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre. The radius is 5 cm. Find the distance between their midpoints.
For chord 6 cm:
Half chord = 3 cm.
Distance from centre:
= √(5^2 - 3^2)
= √(25 - 9)
= 4 { cm}
For chord 8 cm:
Half chord = 4 cm.
Distance from centre:
= √(5^2 - 4^2)
= √(25 - 16)
= 3 { cm}
Since chords are on opposite sides of the centre:
{Distance between midpoints} = 4 + 3 = 7 { cm}
Q2. Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre. The distance between them is 7 cm. Find the radius.
Let the radius be (r).
For chord 10 cm:
Half chord = 5 cm.
Distance from centre:
= √(r^2 - 25)
For chord 24 cm:
Half chord = 12 cm.
Distance from centre:
= √(r^2 - 144)
The longer chord is closer to the centre.
So:
√(r^2 - 25) - √(r^2 - 144) = 7
Try (r = 13).
For 10 cm chord:
√(169 - 25) = √(144) = 12
For 24 cm chord:
√(169 - 144) = √(25) = 5
Difference:
12 - 5 = 7
So, radius:
13 { cm}
Q3. In a circle with centre O, the central angle AOB is 60°. If radius is 12 cm, find chord AB.
Since (OA = OB = 12) cm and (∠ AOB = 60^°), (△ AOB) is equilateral.
So:
AB = 12 { cm}
Q4. If (MN) is a diameter of a circle and (P) is a point on the circle, what is (∠ MPN)?
Since (MN) is a diameter, it subtends a right angle at any point on the circle.
So:
∠ MPN = 90^°
Q5. If the exterior angle of a cyclic quadrilateral is given, how is it related to the opposite interior angle?
The exterior angle of a cyclic quadrilateral equals the interior opposite angle.
This follows because the interior angle and exterior angle form a linear pair, while opposite angles of a cyclic quadrilateral add to (180^°).
Useful Resources for Class 9 Maths