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# Circles Class 9 Notes CBSE Maths Chapter 9 (Free PDF Download)

Last updated date: 11th Sep 2024
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## Class 9 Maths Revision Notes for Circles of Chapter 9 - Free PDF Download

Maths has always been challenging as remembering the formulae is even tougher for a lot. But, with Vedantu’s class 9 Maths Chapter circle notes, we help you remember all of it at a glance. Our notes are designed as per the latest CBSE curriculum helping you understand the concepts quickly and easily. Moreover, the download option for CBSE Class 9 Maths Notes Chapter 9 Circles provides ease of offline study. So, if you want to grasp the concepts quickly or require hands-on formulae while studying, Vedantu got you covered. You can also download Class 9 Science Solutions and CBSE Solutions from Vedantu.com.

Also, check CBSE Class 9 Maths revision notes for all chapters:

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## Introduction

### Circle:

• The locus of the points at a certain distance from a fixed point is defined as a circle.

### Chord:

• A chord is a straight line that connects any two points on a circle.

• A chord is represented by the letters $\text{AB}$ .

• If the longest chord passes  through the centre of the circle, it is termed as the diameter.

• The radius is twice as long as the diameter.

• A diameter is referred to as a $\text{CD}$.

• A secant is a line that divides a circle in half.

• $\text{PQR}$ is a secant of a circle.

### Circumference:

• Circumference refers to the length of a full circle.

• The circumference of a circle is defined as the border curve (or perimeter) of the circle.

### Arc:

• An arc is any section or a part of the circumference.

• A diameter divides a circle into two equal pieces.

• A minor arc is one that is smaller than a semicircle.

• A major arc is one that is larger than a semicircle.

• $\overset\frown{\text{ADC}}$ is a minor arc, whereas $\overset\frown{\text{ABC}}$ is a major arc.

### Sector:

• A sector is the area between an arc and the two radii that connect the arc's centre and end points.

• A segment is a section of a circle that has been cut off by a chord.

### Concentric Circles:

Concentric circles are circles with the same centre.

### Theorem $\text{1}$:

A straight line drawn from the centre of a circle to bisect a chord which is not a diameter, is at right angles to the chord.

• Given Data:

• Here, $\text{AB}$ is a chord of a circle with the centre $\text{O}$.

• The midpoint of $\text{AB}$ is $\text{M}$.

• $\text{OM}$ is joined.

• To Prove:

$\angle \text{AMO = }\angle \text{BMO = 9}{{\text{0}}^{\text{0}}}$

• Construction:

Join $\text{AO}$ and $\text{BO}$.

• Proof:

In $\Delta \text{AOM }$and $\Delta \text{BOM }$

 Statement Reason $\text{AO = BO}$ radii $\text{AM = BM}$ Data $\text{OM = OM}$ Common $\Delta \text{AOM }\cong \Delta \text{BOM }$ $\left( \text{S}\text{.S}\text{.S} \right)$ $\therefore \angle \text{AMO = }\angle \text{BMO}$ Statement $\left( \text{4} \right)$ But $\angle \text{AMO + }\angle \text{BMO = 18}{{\text{0}}^{\text{0}}}$ Linear pair $\therefore \angle \text{AMO = }\angle \text{BMO = 9}{{\text{0}}^{\text{0}}}$ Statements $\left( 5 \right)$ and  $\left( 6 \right)$

Theorem $2$ (Converse of theorem $1$):

The perpendicular to a chord from the centre of a circle bisects the chord.

• Given Data:

1. Here, $\text{AB}$ is a chord of a circle with the centre $\text{O}$.

2. $\text{OM}\bot \text{AB}$

• To Prove:

$\text{AM = BM}$

• Construction:

Join $\text{AO}$ and $\text{BO}$.

• Proof:

In $\Delta \text{AOM }$and $\Delta \text{BOM }$

 Statement Reason $\text{AMO = }\angle \text{BMO }$ Each $\text{9}{{\text{0}}^{0}}$ (data) $\text{AO = BO}$ Radii $\text{OM = OM}$ Common $\Delta \text{AOM }\cong \Delta \text{BOM }$ $\left( \text{R}\text{.H}\text{.S} \right)$ $\text{AM = BM}$ Statement $\left( \text{4} \right)$

The transposition of a statement consisting of 'data' and 'to prove' is the converse of a theorem.

We can see how it works by looking at the previous two theorems:

 Theorem Converse of Theorem $\text{1}$ Data: $\text{M}$ is the midpoint of $\text{AB}$ To Prove: $\text{M}$ is the midpoint of $\text{AB}$ $\text{2}$ To Prove: $\text{OM}\bot \text{AB}$ Data: $\text{OM}\bot \text{AB}$

Theorem $3$:

Equal chords of a circle are equidistant from the centre.

• Given Data:

• Here, $\text{AB}$and $\text{CD}$are equal chords of a circle with centre $\text{O}$.

• $\text{OK}\bot \text{AB}$ and $\text{OL}\bot \text{CD}$

• To Prove:

$\text{OK = OL}$

• Construction:

Join $\text{AO}$ and $\text{CO}$.

• Proof:

 Statement Reason $\text{AK = }\dfrac{1}{2}\text{ AB}$ $\bot$ from the centre bisects the chord. $\text{CL = }\dfrac{1}{2}\text{ CD}$ $\bot$ from the centre bisects the chord. But $\text{AB = CD}$ data $\therefore \text{AK = CL }$ Statements $\left( 1 \right),\left( 2 \right)$ and  $\left( 3 \right)$ In $\Delta \text{AOK}$and $\Delta \text{COL }$ $\angle \text{AKO = }\angle \text{CLO}$ Each $\text{9}{{\text{0}}^{0}}$ (data) $\text{AO = CO}$ radii $\text{AK = CL }$ Statements $\left( 4 \right)$ $\therefore \Delta \text{AOK }\cong \Delta \text{COL}$ $\left( \text{R}\text{.H}\text{.S} \right)$ $\therefore \text{OK = OL }$ Statements $\left( 8 \right)$

Theorem $4$(Converse of theorem $3$):

Chords which are equidistant from the centre of a circle are equal.

• Given Data:

• Here, $\text{AB}$and $\text{CD}$are equal chords of a circle with centre $\text{O}$.

• $\text{OK}\bot \text{AB}$ and $\text{OL}\bot \text{CD}$

• $\text{OK = OL}$

• To Prove:

$\text{AB = CD}$

• Construction:

Join $\text{AO}$ and $\text{CO}$.

• Proof:

In $\Delta \text{AOK}$and $\Delta \text{COL }$

 Statement Reason $\angle \text{AKO = }\angle \text{CLO}$ Each $\text{9}{{\text{0}}^{0}}$ (data) $\text{AO = CO}$ radii $\text{OK = OL }$ data $\Delta \text{AOK }\cong \Delta \text{COL}$ $\left( \text{R}\text{.H}\text{.S} \right)$ $\therefore \text{AK = CL }$ Statements $\left( 4 \right)$ But $\text{AK = }\dfrac{1}{2}\text{ AB}$ $\bot$ from the centre bisects the chord. $\text{CL = }\dfrac{1}{2}\text{ CD}$ $\bot$ from the centre bisects the chord. $\therefore \text{AB = CD}$ Statements $\left( 5 \right),\left( 6 \right)$ and  $\left( 7 \right)$

Theorem $5$:

There is one circle, and only one, which passes through three given points not in a straight line.

• Given Data:

Here, $\text{X, Y}$and $\text{Z}$ are three points not in a straight line.

• To Prove:

A unique circle passes through $\text{X, Y}$and $\text{Z}$.

• Construction:

• Join $\text{XY}$ and $\text{YZ}$.

• Draw perpendicular bisectors of $\text{XY}$ and $\text{YZ}$ to meet at $\text{O}$.

• Proof:

 Statement Reason $\text{OX = OY}$ $\text{O}$ lies on the $\bot$ bisector of $\text{XY}$ $\text{OY = OZ}$ $\text{O}$ lies on the $\bot$ bisector of $\text{YZ}$ $\text{OX = OY = OZ}$ Statements $\left( 1 \right)$ and $\left( 2 \right)$ $\text{O}$ is the only point equidistant from $\text{X, Y}$and $\text{Z}$. Statements $\left( 3 \right)$ With $\text{O}$ as centre and radius $\text{OX}$, a circle can be drawn to pass through $\text{X, Y}$and $\text{Z}$. Statements $\left( 4 \right)$ Therefore, the circle with centre $\text{O}$ is a unique circle passing through $\text{X, Y}$and $\text{Z}$. Statements $\left( 5 \right)$

### Angle Properties (Angle, Cyclic Quadrilaterals and Arcs):

• In figure$\left( \text{i} \right)$ , the straight line $\text{AB}$students $\angle \text{APB}$ on the circumference.

$\angle \text{APB}$can be said to be subtended by arc $\text{AMB}$, on the remaining part of the circumference.

• In fig. $\left( \text{ii} \right)$ , arc $\text{AMB}$ subtends $\angle \text{APB}$ on the circumference, and it subtends $\angle \text{AOB}$ at the centre.

• In fig. $\left( \text{iii} \right)$, $\angle \text{APB}$ and $\angle \text{AQB}$ are in the same segment.

• Now we will go through the theorems based on the angle properties of the circles.

Theorem $6$:

The angle at which an arc of a circle subtends at the centre is double the angle which it subtends at any point on the remaining part of the circumference.

• Given Data:

Arc $\text{AMB}$subtends $\square \text{AOB}$ at the center $\text{O}$ of the circle and $\square \text{APB}$ on the remaining part of circumference.

• To Prove:

$\angle \text{AOB = 2}\angle \text{APB}$

• Construction:

Join $\text{PO}$ and produce it to $\text{Q}$.

Let, $\angle \text{APQ = x}$ and $\angle \text{BPQ = y}$

• Proof:

 Statement Reason $\angle \text{AOQ = }\angle \text{x + }\angle \text{A}$ $\text{Ext}\text{. }\angle =\text{sum of the int}\text{. opp}\text{. }\angle \text{s}$ $\angle \text{x = }\angle \text{A}$ $\because \text{ OA = OP}$ (Radii) $\therefore \angle \text{AOQ = 2}\angle \text{x}$ Statements $\left( 1 \right)$ and $\left( 2 \right)$ $\therefore \angle \text{BOQ = 2}\angle \text{y}$ Same way as Statements $\left( 3 \right)$ From figure $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ $\angle \text{AOQ + }\angle \text{BOQ = 2}\angle \text{x + 2}\angle \text{y}$ Statements $\left( 3 \right)$ and $\left( 4 \right)$ $\Rightarrow \text{ }\angle \text{AOB = 2}\left( \angle \text{x + }\angle \text{y} \right)$ Statements $\left( 5 \right)$ From figure $\left( \text{ii} \right)$ $\angle \text{BOQ}-\angle \text{AOQ = 2}\angle \text{y}-\text{2}\angle \text{x}$ Statements $\left( 3 \right)$ and $\left( 4 \right)$ $\angle \text{AOB = 2}\left( \angle \text{y - }\angle \text{x} \right)$ Statements $\left( 8 \right)$ $\because \angle \text{AOB = 2}\angle \text{APB}$ Statements $\left( 9 \right)$

Theorem $7$:

Angles in the same segment of a circle are equal.

• Given Data:

$\angle \text{APB}$ and $\angle \text{AQB}$ are in the same segment of a circle with center $\text{O}$.

• To Prove:

$\angle \text{APB = }\angle \text{AQB}$

• Construction:

Join $\text{AO}$ and $\text{BO}$.

Let arc $\text{AMB}$subtend angle $\text{x}$ at the center $\text{O}$.

• Proof:

 Statement Reason $\angle \text{x = 2}\angle \text{APB}$ $\angle \text{at center}=2\times \angle \text{on the circumference}$ $\angle \text{x = 2}\angle \text{AQB}$ $\angle \text{at center}=2\times \angle \text{on the circumference}$ $\therefore \angle \text{APB = }\angle \text{AQB}$ Statements $\left( 1 \right)$ and $\left( 2 \right)$

Theorem $8$:

The angle in a semicircle is a right angle.

• Given Data:

$\text{AB}$ is the diameter of a circle with center $\text{O}$.

$\text{P}$ is any point on the circle.

• To Prove:

$\angle \text{APB = }{{90}^{\circ }}$

• Proof:

 Statement Reason $\angle \text{APB = }\dfrac{1}{2}\angle \text{AOB}$ $\angle \text{at center}=2\times \angle \text{on the circumference}$ $\angle \text{AOB = 18}{{\text{0}}^{\circ }}$ $\text{AOB}$ is a straight line. $\therefore \angle \text{APB = }\dfrac{1}{2}\times {{180}^{\circ }}$ Statements $\left( 1 \right)$ and $\left( 2 \right)$ $\therefore \angle \text{APB = }{{90}^{\circ }}$ Statements $\left( 3 \right)$.

• If the vertices of a quadrilateral lie on a circle, the quadrilateral is called a cyclic quadrilateral.

• The vertices are known as concyclic points.

• From the above figure, $\text{ABCD}$ is a cyclic quadrilateral.

The vertices $\text{A, B, C}$ and $\text{D}$ are concyclic points.

Theorem $9$:

The opposite angles of a quadrilateral inscribed in a circle (cyclic) are supplementary.

• Given Data:

$\text{ABCD}$ is a cyclic quadrilateral.

$\text{O}$ is the center  of a circle.

• To Prove:

1. $\angle \text{A+}\angle \text{C = 18}{{0}^{\circ }}$

2. $\angle \text{B+}\angle \text{D = 18}{{0}^{\circ }}$

• Proof:

 Statement Reason $\angle \text{A = }\dfrac{1}{2}\angle \text{x}$ $\angle \text{at center}=2\times \angle \text{on the circumference}$ $\angle \text{C = }\dfrac{1}{2}\angle \text{y}$ $\angle \text{at center}=2\times \angle \text{on the circumference}$ $\angle \text{A + }\angle \text{C = }\dfrac{1}{2}\angle \text{x + }\dfrac{1}{2}\angle \text{y}$ Statements $\left( 1 \right)$ and $\left( 2 \right)$ $\angle \text{A + }\angle \text{C = }\dfrac{1}{2}\left( \angle \text{x + }\angle \text{y} \right)$ Statements $\left( 3 \right)$ But $\angle \text{x + }\angle \text{y}={{360}^{\circ }}$ $\angle$ at a point $\therefore \angle \text{A + }\angle \text{C = }\dfrac{1}{2}\times {{360}^{\circ }}$ Statements $\left( 4 \right)$ and $\left( 5 \right)$ $\therefore \angle \text{A + }\angle \text{C = }{{180}^{\circ }}$ Statements $\left( 6 \right)$ Also, $\angle \text{ABC + }\angle \text{ADC = }{{180}^{\circ }}$ Same way as statements $\left( 7 \right)$

### Corollary:

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

• Given Data:

$\text{ABCD}$ is a cyclic quadrilateral.

$\text{BC}$ is produced to $\text{E}$

• To Prove:

$\angle \text{DCE}=\angle \text{A}$

• Proof:

 Statement Reason $\angle \text{A +}\angle \text{BCD}={{180}^{\circ }}$ Opp. $\angle \text{s}$ of a cyclic quad. $\angle \text{BCD +}\angle \text{DCE}={{180}^{\circ }}$ Linear pair $\therefore \angle \text{BCD +}\angle \text{DCE}=\angle \text{A}+\angle \text{BCD}$ Statements $\left( 1 \right)$ and $\left( 2 \right)$ $\therefore \angle \text{DCE}=\angle \text{A}$ Statements $\left( 2 \right)$

### Alternate Segment Property

Theorem $10$:

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

• Given Data:

A straight line $\text{SAT}$ touches a given circle with centre $\text{O}$ at $\text{A}$. $\text{AC}$ is a chord through the point of contact $\text{A}$.

$\angle \text{ADC}$ is an angle in the alternate segment to $\angle \text{CAT}$ and $\angle \text{AEC}$is an angle in the alternate segment to $\angle \text{CAS}$

• To Prove:

1. $\angle \text{CAT}=\angle \text{ADC}$

2. $\angle \text{CAS}=\angle \text{AEC}$

• Construction:

Draw $\text{AOB}$ as diameter and join $\text{BC}$ and $\text{OC}$.

• Proof:

 Statement Reason $\angle \text{OAC = }\angle \text{OCA}=\text{x}$ Since, $\text{OA = OC}$ and supposition $\angle \text{CAT +}\angle \text{x}={{90}^{\circ }}$ Since, tangent-radius property $\angle \text{AOC +}\angle \text{x + }\angle \text{y}={{180}^{\circ }}$ Sum of angles of a triangle $\angle \text{AOC}={{180}^{\circ }}-2\angle \text{x}$ Statements $\left( 3 \right)$ Also, $\angle \text{AOC}=2\angle \text{ADC}$ $\angle \text{at the center}=2\angle \text{on the circle}$ $\angle \text{CAT}={{90}^{\circ }}-\text{x}$ Statements $\left( 2 \right)$ $2\angle \text{CAT}={{180}^{\circ }}-2\text{x}$ Statements $\left( 6 \right)$ $\therefore 2\angle \text{CAT}=2\angle \text{ADC}$ Statements $\left( 4 \right)$, $\left( 5 \right)$ and $\left( 7 \right)$ $\angle \text{CAT}=\angle \text{ADC}$ Statements $\left( 8 \right)$ $\angle \text{CAS +}\angle \text{CAT}={{180}^{\circ }}$ Linear pair $\angle \text{ADC +}\angle \text{AEC}={{180}^{\circ }}$ Opp. Angles of a cyclic quad $\angle \text{CAS +}\angle \text{CAT}=\angle \text{ADC +}\angle \text{AEC}$ Statements $\left( 10 \right)$ and $\left( 11 \right)$ $\therefore \angle \text{CAS}=\angle \text{AEC}$ Statements $\left( 9 \right)$ and $\left( 12 \right)$

Theorem $11$:

In equal circles (or in the same circle), if two arcs subtend equal angles at the centres, they are equal.

• Given Data:

$\text{AXB}$ and $\text{CYD}$ are equal circles with centers $\text{P}$ and $\text{O}$.

Arcs $\text{AMD, CND}$ subtend equal angles $\text{APB, CQD}$.

• To Prove:

$\text{arc AMD = arc CND}$

• Proof:

 Statement Reason Apply $\bigcirc \text{ CYD}$ to $\bigcirc \text{ AXB}$ so that center $\text{Q}$ falls on center $\text{P}$ and $\text{QC}$ along $\text{PA}$ and $\text{D}$ on the same side as $\text{B}$.Therefore, $\bigcirc \text{ CYD}$ overlaps $\bigcirc \text{ AXB}$ Since, circles are equal (data) $\therefore \text{C}$ falls on $\text{A}$ Since, $\text{PA = QC}$ (data) $\angle \text{APB = }\angle \text{CQD}$ data $\therefore \text{QD}$ falls along $\text{PB}$ Statements $\left( 1 \right)$ and $\left( 3 \right)$ $\therefore \text{D}$ falls on $\text{B}$ Since, $\text{QD = PB}$ (data) $\therefore \text{ arc CND}$ coincides with $\text{arc AMD}$ Statements $\left( 2 \right)$ and $\left( 5 \right)$ $\text{arc AMD = arc CND}$ Statements $\left( 6 \right)$

Theorem $12$ (Converse of $11$):

In equal circles (or in the same circle) if two arcs are equal, they subtend equal angles at the centres.

• Given Data:

In equal circles $\text{AXB}$ and $\text{CYD}$, equal arcs $\text{AMD}$ and $\text{CND}$ subtend $\angle \text{APB}$ and $\angle \text{CQD}$ at the centers $\text{P}$ and $\text{Q}$ respectively.

• To Prove:

$\angle \text{APB = }\angle \text{CQD}$

• Proof:

 Statement Reason Apply $\bigcirc \text{ CYD}$ to $\bigcirc \text{ AXB}$ so that center $\text{Q}$ falls on center $\text{P}$ and $\text{QC}$ along $\text{PA}$ and $\text{D}$ on the same side as $\text{B}$.Therefore, $\bigcirc \text{ CYD}$ overlaps $\bigcirc \text{ AXB}$ Since, circles are equal (data) $\therefore \text{C}$ falls on $\text{A}$ Since, $\text{PA = QC}$ (data) $\text{arc AMD = arc CND}$ data $\therefore \text{D}$ falls on $\text{B}$ Statements $\left( 1 \right),\left( 2 \right)$ and $\left( 3 \right)$ $\therefore \text{QD}$ coincides with $\text{PB}$and $\text{QC}$ coincides with $\text{PA}$ Statements $\left( 1 \right),\left( 2 \right)$ and $\left( 4 \right)$ $\angle \text{APB = }\angle \text{CQD}$ Statements $\left( 5 \right)$

In case of the same circle:

Figures $\left( \text{ii} \right)$ and $\left( \text{iii} \right)$ can be considered to be two equal circles which are obtained from figure $\left( \text{i} \right)$ and then the above proofs may be applied.

Theorem $13$):

In equal circles (or in the same circle), if two chords are equal, they cut off equal arcs.

• Given Data:

In equal circles $\text{AXB, CYD}$ with centers $\text{P}$ and $\text{Q}$ have

$\text{chord AB = chord CD}$

• To Prove:

$\text{arc AMB = arc CND}$

$\text{arc AXB = arc CYD}$

• Proof:

In $\Delta \text{ABP}$ and $\Delta \text{CDQ}$

 Statement Reason $\text{AP = CQ}$ Radii of equal circles. $\text{BP = DQ}$ Radii of equal circles. $\text{AB = CD}$ Radii of equal circles. $\Delta \text{ABP }\cong \text{ }\Delta \text{CDQ}$ $\left( \text{S}\text{.S}\text{.S} \right)$ $\therefore \angle \text{APB = CQD}$ Statements $\left( 4 \right)$ $\text{arc AMB = arc CND}$ Statements $\left( 5 \right)$ $\bigcirc \text{AXB - arc AMB = }\bigcirc \text{CYD - arc CND}$ Equal arcs [Statements 6] $\therefore \text{ arc AXB = arc CYD}$ Statements $\left( 7 \right)$

Theorem $14$ (Converse of $13$):

In equal circles (or in the same circle) if two arcs are equal, the chords of the arcs are equal.

• Given Data:

Equal circles $\text{AXB, CYD}$ with centers $\text{P}$ and $\text{Q}$ have

$\text{arc AMB = arc CND}$

• To Prove:

$\text{chord AB = chord CD}$

• Construction:

Join $\text{AP, BP, CQ}$ and $\text{DQ}$.

• Proof:

In $\Delta \text{ABP}$ and $\Delta \text{CDQ}$

 Statement Reason $\text{AP = CQ}$ Radii of equal circles. $\text{BP = DQ}$ Radii of equal circles. $\angle \text{APB = CQD}$ $\because \text{ arc AMB = arc CND}$ $\therefore \Delta \text{ABP }\cong \text{ }\Delta \text{CDQ}$ $\left( \text{S}\text{.A}\text{.S} \right)$ $\therefore \text{AB = CD}$ Statements $\left( 4 \right)$

Theorem $15$:

If two chords of a circle intersect internally, then the product of the length of the segments are equal.

• Given Data:

$\text{AB}$ and $\text{CD}$ are chords of a circle intersecting externally at $\text{P}$.

• To Prove:

$\text{AP }\!\!\times\!\!\text{ BP = CP}\times \text{DP}$

• Construction:

Join $\text{AC}$ and $\text{BD}$.

• Proof:

In $\Delta \text{APC}$ and $\Delta \text{DPB}$

 Statement Reason $\angle \text{A = }\angle \text{D}$ Angles in the same segment. $\angle \text{C = }\angle \text{B}$ Angles in the same segment. $\therefore \text{ }\Delta \text{APC }\!\!\sim\Delta \text{DPB}$ $\text{AA}$ similarity $\therefore \dfrac{\text{AP}}{\text{DP}}\text{=}\dfrac{\text{CP}}{\text{BP}}$ Statements $\left( 3 \right)$ $\therefore \text{AP}\times \text{BP = CP }\!\!\times\!\!\text{ DP}$ Statements $\left( 4 \right)$

Theorem $16$:

If two chords of a circle intersect externally, then the product of the lengths of the segments are equal.

• Given Data:

$\text{AB}$ and $\text{CD}$ are chords of a circle intersecting externally at $\text{P}$.

• To Prove:

$\text{AP }\!\!\times\!\!\text{ BP = CP}\times \text{DP}$

• Construction:

Join $\text{AC}$ and $\text{BD}$.

• Proof:

In $\Delta \text{ACP}$ and $\Delta \text{DBP}$

 Statement Reason $\angle \text{A = }\angle \text{BDP}$ $\text{Ext}\text{. }\angle \text{of a cyclic quad}\text{. = Int}\text{. opp}\text{. }\angle$ $\angle \text{C = }\angle \text{DBP}$ $\text{Ext}\text{. }\angle \text{of a cyclic quad}\text{. = Int}\text{. opp}\text{. }\angle$ $\therefore \text{ }\Delta \text{ACP }\!\!\sim\Delta \text{DBP}$ $\text{AA}$ similarity $\therefore \dfrac{\text{AP}}{\text{DP}}\text{=}\dfrac{\text{CP}}{\text{BP}}$ Statements $\left( 3 \right)$ $\text{AP}\times \text{BP = CP }\!\!\times\!\!\text{ DP}$ Statements $\left( 4 \right)$

Theorem $17$:

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square on the length of the tangent from the point of contact to the point of intersection.

• Given Data:

A chord $\text{AB}$ and a tangent $\text{TP}$ at a point $\text{T}$ on the circle intersect at $\text{P}$.

• To Prove:

$\text{AP }\!\!\times\!\!\text{ BP = P}{{\text{T}}^{\text{2}}}$

• Construction:

Join $\text{AT}$ and $\text{BT}$.

• Proof:

 Statement Reason In $\Delta \text{APT}$ and $\Delta \text{TPB}$ Angles in alternate segment $\angle \text{A = }\angle \text{BTP}$ $\angle \text{P = }\angle \text{P}$ Common $\therefore \text{ }\Delta \text{APT }\!\!\sim\Delta \text{TPB}$ $\text{AA}$ similarity $\dfrac{\text{AP}}{\text{PT}}\text{=}\dfrac{\text{PT}}{\text{BP}}$ Statements $\left( 3 \right)$ $\text{AP}\times \text{BP = P}{{\text{T}}^{2}}$ Statements $\left( 4 \right)$

### Test for Concyclic Points:

1. Conversely, the statement, 'Angles in the same segment of a circle are equal', is one test for concyclic points.

We state:

If two equal angles are on the same side of a line and are subtended by it, then the four points are concyclic.

In the figure, if $\angle P=\angle Q$and the points $P,Q$ are on the same side of $AB$, then the points $A,B,P$ and $Q$are concyclic.

1. Converse of 'opposite angles of a cyclic quadrilateral are supplementary' is one more test for concyclic points.

We state:

If the opposite angles of a quadrilateral are supplementary, then its vertices are concyclic.

In the figure, if $\angle \text{A}+\angle C={{180}^{\circ }}$ then $\text{A, B, C}$ and $\text{D}$ are concyclic points.

## Key Takeaways of NCERT Solutions Class 9 Maths Chapter 9 - Circles Free PDF

The circles class 9 questions with solutions allow the students to take several benefits. They are:

• Students can score better marks and gain good knowledge.

• Students can take either a soft copy or hard copy.

## Conclusion

NCERT Solutions for Class 9 Maths Chapter 9 Circles is a valuable resource for students who want to gain a sound knowledge in logical thinking and problem solving. It provides students with a clear and concise explanation of the concepts, as well as a large number of solved and unsolved problems. Students can download the solutions for free from the official website, which makes it a convenient and affordable resource. In addition to NCERT Solutions for Class 9 Maths Chapter 9 Circles, Vedantu also offers NCERT Solutions for Class 9 Science and CBSE Solutions for all subjects. These solutions are up-to-date and are sure to help students in their academic journey.

## FAQs on Circles Class 9 Notes CBSE Maths Chapter 9 (Free PDF Download)

1. What is a Cyclic Quadrilateral and What is the Theorem Associated with it?

When all the quadrilateral vertices lie on a circle, they are defined as a cyclic quadrilateral.

According to the theorem, any pair of the opposite angles in a cyclic quadrilateral will have the sum of 180°.

The reverse of this theorem states that if the opposite angles in a quadrilateral have a sum of 180°, then the quadrilateral is cyclic.

2. What Happens When the Angles are Subtended From the Common Chord?

The angles subtended from the common chord and which lie on the same segment are always equal.

3. What Happens When an Angle is Subtended at the Centre and Another Angle is Subtended by the Same Arc on Another Point?

According to a theorem, an angle subtended by the arc at the centre is twice the angle that is subtended by the same arc on any other point of the circle.

4. How are NCERT Solutions for Class 9 Maths Chapter 9 helpful for students?

One of the best Solutions out there is provided by Vedantu and their Solutions for chapter 9 of Class 9 Maths helps students in a thorough understanding of the chapter through an in-depth explanation of concepts in short, precise and simple expert-created answers. The circle is a chapter that requires proper comprehension and forms a base during Class 9 for future classes. Therefore, the student must go through all the solutions available free of cost to ensure they do well.

5. Why should we follow NCERT Solutions for Class 9 Maths Chapter 9?

NCERT Solutions offer many benefits to students. These Solutions help students use their time in a meticulous manner by offering them factual analysis and thorough explanation on the concepts mentioned in Chapter 9 Of Class 9 Mathematics in an accurate and short display of information. The structure of the language used is easy to follow, thereby allowing students to easily grasp the concepts. Students can download these Solutions from the vedantu website (vedantu.com) or the vedantu app, and be rest assured with their exams.

6. What does chapter 9 of class 9 Maths cover?

In class 9, students learn Circles in chapter 9 of the NCERT textbook. Revision Notes are good references to learn and revise about the different concepts taught in the chapter like

• Circles

• Angle Subtended by a chord

• Equal Chords with distances from the centre

• Angle Subtended by an arc

The chapter also covers some major and important theorems that students need to memorize and then apply for solving problems of circles, quadrilaterals, triangles, etc. from the chapter.

7. What are the important theorems in chapter 9 of class 9 Maths?

Chapter 9 of Circles for class 9 Maths has some major theorems that students need to thoroughly learn and memorize to solve all the questions thoroughly. There are 5 major theorems and their converses and the following points are an attempt at stating the theorems in brief:

• Equal chords subtending angles at the centre

• Perpendicular to a chord

• Equal chords are equidistant from the centre

• The measure of angles subtended by the same arc

• Opposite angles are supplementary in quadrilaterals

8. What are the benefits of using revision notes for learning the chapter of circles in Class 9?

Revision notes are very beneficial once students learn how these notes help them with their exam preparations. For chapter 9 of class 9 Maths, which deals with circles, their concepts, and various theorems, things can get very complicated and confusing for the students. The notes offered by Vedantu For Class 9 Maths are one of the best because they provide detailed, but short notes which simplify all these complicated concepts, thereby providing students with an easy learning experience and last-minute help. Also these revision notes are available free of cost to download.