CBSE Notes Class 9 Maths Chapter 9 Circle

In Class 9 Maths, Chapter 10 deals with the fascinating and fundamental topic of circles. A circle, one of the most basic shapes in geometry, is pivotal in understanding more complex geometric concepts. This CBSE Class 9 Maths Chapter provides students with a foundational understanding of circles, their properties, and key theorems.

1.0Download CBSE Notes for Class 9 Maths Chapter 9 Circle - Free PDF!!

Unlock your understanding of CBSE Class 9 Maths Chapter 9 – Circle with our free, easy-to-download PDF notes. These CBSE notes are designed to help students grasp the key concepts of circles, including properties, theorems, and important formulas.

2.0Definition and Basic Concepts 

A circle is a set of all points in a plane that are at a constant distance (radius) from a given point (center). Here are some essential terms related to circles:

• Center: The fixed point from which every point on the circle is equidistant. (Represented as O)
• Radius: The straight-line distance from the center of the circle to any point on its circumference. (Represented as OC)
• Diameter: The distance across the circle through the center, which is twice the radius. (Represented as AOB)

• Chord: A line segment with both of its endpoints lying on the circle's boundary. The diameter is a special chord that passes through the center. (Represented as PQ)

• Arc: A part of the circumference of a circle. (Represented minor arc as , and major arc as )

• Sector: The area bounded by two radii and the arc between them. (Represented as POQ)

• Segment: The region enclosed by a chord and the arc it subtends.

3.0Theorems Related to Circles

The CBSE Class 9 Maths Notes Chapter 9 Circle covers several important theorems related to circles:

Theorem 1: Chords of the same length in a circle form equal angles at the center.

Theorem 2: If two chords in a circle subtend equal angles at the center, then the chords are of equal length.

Theorem 3: A perpendicular drawn from the center of a circle to a chord divides the chord into two equal parts.

Theorem 4: A line passing through the center of a circle that divides a chord into two equal parts is perpendicular to the chord.

Theorem 5: A single, unique circle can be drawn through any three given non-collinear points. (Here A, B, and C are non- collinear points)

Theorem 6: Chords of equal length in a circle (or in congruent circles) are equally distant from the center (or centers).

Theorem 7: Chords that are equally distant from the center of a circle are of equal length.

Theorem 8: The angle formed at the center by an arc is twice as large as the angle formed by the same arc at any point on the rest of the circle.

Theorem 9: Angles that lie within the same segment of a circle are equal/same.

Theorem 10: If a line segment joining two points creates equal angles at two other points located on the same side of the line containing the segment, then all four points lie on a common circle (i.e., they are concyclic).

Cyclic Quadrilaterals

Theorem 11: In a cyclic quadrilateral, the sum of the opposite angles always adds up to 180º.

∠A + ∠C = 180° and ∠B + ∠D = 180°

Theorem 12: A quadrilateral is cyclic if the sum of each pair of its opposite angles equals 180º.

4.0Solved Example on Circles

Example 1: In figure O is the centre of the circle of radius 15 cm. OP ⊥ AB, OQ ⊥ CD, AB || CD, AB = 18 cm and CD = 22 cm. Find PQ. 

Solution:

Given, AB and CD are two parallel chords of a circle with center O. 

Also, OP ⊥ AB and OQ ⊥ CD. 

Since AB || CD, so points O, P, and Q are collinear. 

Join OA and OC. 

We have AB = 18 cm, CD = 22 cm, and OA = 15 cm [Radii of the circle]

Since the perpendicular drawn from the center of the circle to a chord bisects it, therefore:

$\mathrm{AP}=\frac{1}{2}\mathrm{AB}$

$\Rightarrow \mathrm{AP}=\frac{1}{2}\times 18$

$\Rightarrow AP=9\mathrm{cm}.$

And C Q= $\frac{1}{2}CD$

$\Rightarrow \mathrm{CQ}=\frac{1}{2}\times 22$

$\Rightarrow CQ=11\mathrm{cm}.$

Now, in the right triangle AOP.

OP² = OA² – AP² 

⇒ OP² = 15² – 9² 

⇒ OP = $\sqrt{225-81}$

⇒ OP = $\sqrt{144}$

⇒ OP = 12

In the right triangle COQ.

OQ² = OC² – CQ² 

⇒ OQ² = 15² – 11² 

⇒ OQ = $\sqrt{225-121}$

⇒ OQ = $\sqrt{104}$

⇒ OQ = $2\sqrt{26}\mathrm{cm}$

Now, PQ = OP – OQ

Given 

PQ = 12 – $2\sqrt{26}\mathrm{cm}$

P Q= $2(6-\sqrt{26})\mathrm{cm}$

Example 2: AB and CD are two parallel chords of a circle with lengths 24 cm and 10 cm respectively, lying on the same side of its center O. If the distance between the chords is 7 cm, find the radius of the circle. 

Solution:

Draw OL ⊥ AB at L and OM ⊥ CD at M. 

Since AB || CD, it follows that O, L, and M are collinear.

Now, OL ⊥ AB and OM ⊥ CD. 

$\Rightarrow AL=\frac{1}{2}AB=\frac{1}{2}\times 24=12\mathrm{cm}$

And C M= $\frac{1}{2}CD=\frac{1}{2}\times 10=5\mathrm{cm}$

Also, the distance between the chords = LM = 7 cm.

Join OA and OC.

Let OL = x cm. Then, OM = (x + 7) cm.

Let the radius of the circle be r cm.

Then, OA = OC = r cm.

Example 3: If Bisector AD of ∠BAC of ΔABC passes through the centre O of the circumcircle of ΔABC. If AB = 5 cm, find AC.

Solution: 

Given: Bisector AD of ∠BAC of ΔABC passes through the centre O of the circumcircle of ΔABC. 

Construction: Draw OP ⊥ AB and OQ ⊥ AC

In ∆APO and ∆AQO,

∠OAP = ∠OAQ (Given)

∠OPA = ∠OQA (Each = 90° by construction)

OA = OA (Common)

Therefore, by AAS,

∆APO ≅ ∆AQO

So:

OP = OQ

OP = OQ (By C.P.C.T.)

AB = AC (∴ Chords equidistant from the centre are equal)

∴ AC = AB = 5 cm

Then, OA = OC = r cm.

Now, from right- angled triangles ΔOLA and ΔOMC, we have

$O{A}^2=O{L}^2+A{L}^2\text{ and }O{C}^2=O{M}^2+C{M}^2$

$=r^2+x^2+(12{)}^2\dots (\mathrm{i})$

And

$r^2=(x+7{)}^2+(5{)}^2\dots \text{ (ii) }$

$\Rightarrow x^2+(12{)}^2=(x+7{)}^2+(5{)}^2$

$\Rightarrow x^2+144x=x^2+14x+74$

$\Rightarrow 14x=70$

$\Rightarrow x=5$

Substituting (x = 5) in equation (i), we get:

$r^2=x^2+(12{)}^2=(5{)}^2+(12{)}^2=(25+144)=169$

Then,

$r=\sqrt{169}=13\mathrm{cm}$

Hence, the radius of the circle is 13 cm.

Example 4: Find the distance of a chord from the center of a circle with a radius of 13 cm. The chord AB is 10 cm long.

Solution:

Draw OL perpendicular to AB from the center O.

Since the perpendicular from the center of a circle to a chord bisects the chord:

$AL=LB=\frac{1}{2}\times AB=\frac{1}{2}\times 10\mathrm{cm}=5\mathrm{cm}$

In the right triangle OLA, we have:

OA² = OL² + AL² 

⇒ 13² = OL² + 5² 

⇒ 169 = OL2 + 25 

⇒ OL² = 144 

$\Rightarrow \mathrm{OL}=\sqrt{144}=12\mathrm{cm}$

Hence, the distance of the chord from the center is 12 cm.

Example 5: Determine the length of a chord that is 5 cm away from the center of a circle with a radius of 13 cm.

Solution: 

Let the chord of a circle be AB with centre O and radius 13 cm. Draw OL ⊥ AB and join OA. Clearly, OL = 5 cm and OA = 13 cm.

In the right triangle OLA, we have…

OA² = OL² + AL²

⇒ 13² = 5² + AL²

⇒ AL² = 144

⇒ AL = 12

Since the perpendicular from the center to a chord bisects the chord, therefore,

AB = 2AL = (2 × 12) cm = 24 cm

Example 6: In below figure, O is the centre of the circle, with a radius of 5 cm. OP ⊥ AB, OQ ⊥ CD, AB || CD and AB = 6 cm and CD = 8cm. Determine PQ.

Solution: 

Join OA and OC. 

Since OP ⊥ AB and OQ ⊥ CD from the center of the circle, they bisect the chords. Therefore, PC = PD and OA = OC = 5 cm (radius of the circle). O and Q are mid-points of AB and CD respectively.

Consequently, AP = PB = $\frac{1}{2}$ A B  = 4 cm and CQ = QD = $\frac{1}{2}$ C D = 4 cm.

In the right triangle OAP, we have:

⇒ OA² + AP² = OP²

⇒ 5² + 4² = OP²

⇒ 25 + 16 = OP²

⇒ OP² = 41

⇒ OP = $\sqrt{41}$

In the right triangle OCQ:

⇒ OC² - CQ² = OQ²

⇒ 5² - 4² = OQ²

⇒ 25 – 16 = OQ²

⇒ OQ = $\sqrt{9}$

⇒ OQ = 3

Therefore, PQ = OP – OQ = OP – OQ = (4 – 3)) cm = 1 cm

Example 7: AB and CD are 2 parallel chords of a circle, with AB = 6 cm and CD = 12 cm. The distance between the two chords is 3 cm. Determine the radius of the circle.

Solution: 

Let AB and CD be two parallel chords of a circle with centre O, where AB = 6 cm and CD = 12 cm. Let the radius of the circle be r cm. Draw OP⊥ AB and OQ ⊥ CD. Since AB || CD, OP⊥ AB and OQ ⊥ CD. Therefore, points o, Q, and P are collinear. 

Clearly, PQ = 3 cm

Let OQ = x cm. Then, Op = (x + 3) cm

Clearly, AP = PB, CQ = QD. Therefore, AP =$\frac{1}{2}$ A B  = $\frac{1}{2}\times 6$= 3 cm, CQ = $\frac{1}{2}CD=\frac{1}{2}\times 12$

= 6 cm. 

In right triangles OAP and OCQ we have:

OA² = OP² + AP² and OC² = OQ² + CQ²

r² = (r + 3)² + 3² and r² = x² + 6²

(r + 3)² + 3² = x² + 6² [on equating the value of r²]

r² + 6x + 9 + 9 = x² + 36

6x = 18

x = 3 cm

Putting the value of x in r² = x² + 6², we get

r² = 3² + 6² 

r² = 9 + 36

r² = 45

r = $\sqrt{45}$ cm 

Hence, the radius of the circle is r = $\sqrt{45}$ , we get

5.0Key Features of CBSE Maths Notes for Class 9 Chapter 9

1. The radius of the circle is 8 cm., and one of its chords is 12 cm long. Determine the distance from the center to the chord.
2. Calculate the length of a chord that is 5 cm away from the center of a circle with a radius of 10 cm.
3. Find the length of a chord that is 4 cm away from the center of a circle with a radius of 6 cm.
4. Two parallel chords, AB and CD, have lengths of 5 cm and 11 cm respectively. Given that the distance between AB and CD is 3 cm. Determine the radius of the circle.