NCERT Solutions Class 9 Maths Chapter 9 - Circles

In your daily life experiences, you might have noticed some objects having a round shape as in wheels of cars or other vehicles, bangles, the surfaces of clocks, certain coins like the one and five rupee coins or buttons of shirts. The mentioned shapes are examples of circles and that is why we shall focus on the topic of circles covered in the Class 9 maths syllabus chapter 9, which covers definition and properties of a circle. This chapter also presents useful theorems one of which states that a perpendicular from the centre of a circle to a chord will divide the chord into two equal parts while the other states that the chords of a circle which are equal in length lie at an equal distance from the centre of the circle. 

In this article, you will get the NCERT Solutions Class 9 Maths Chapter 9 Circle PDF which is made by the best of ALLEN's experts in depth for your ease in solving the questions. In addition to that, these exercises that are necessary for the completion of Class 9 examinations will also help you develop your logical and structural reasoning skills and geometric problem solving skills as well.

1.0Download NCERT Solutions Class 9 Maths Chapter 9 Circles

Download NCERT Solutions for Class 9 Maths Chapter 9 Circles in a free and easy-to-access PDF format. These solutions provide step-by-step explanations for all exercises, including key concepts like the properties of circles, tangents, and angles subtended by chords.

2.0NCERT Solutions Class 9 Maths Chapter 9 - Circles: All Exercises

3.0What Will Students Learn in Chapter 9 - Circles?

• Circles have a definition and fundamental concepts that constitute them.
• Key terms like radius, diameter, chord, arc, and circumference must be understood.
• Important theorems, such as:
• A jump from the center to any chord fails to lose its focus and accomplishes two things, namely, it is perpendicular and it bisects the chord.
• Two chords of equal length subtended from the center of a circle will be at the same distance from the center.
• What approach to take when these particular components, chords, arcs, and even tangents are applied in the context of a problem.
• Use of theorems to establish and justify relationships between figures contained in circles.

4.0NCERT Questions with Solutions for Class 9 Maths Chapter 9 - Detailed Solutions

Exercise : 10.1

• How many tangents can a circle have? Sol. There can be infinitely many tangents to a circle.
• Fill in the blanks : (i) A tangent to a circle intersects it in ..... point (s). (ii) A line intersecting a circle in two points is called a....... (iii) A circle can have parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called Sol. (i) One (ii) Secant (iii) Two (iv) Point of contact.
• A tangent $PQ$ at a point $P$ of a circle of radius 5 cm meets a line through the centre 0 at a point Q so that $\mathrm{OQ}=12\mathrm{cm}$. Length $PQ$ is. (A) 12 cm (B) 13 cm (C) 8.5 cm (D) $\sqrt{119}\mathrm{cm}$ Sol.

• 0 is the centre of the circle. The radius of the circle is 5 cm . $PQ$ is tangent to the circle at $P$. Then $\mathrm{OP}=5\mathrm{cm}$ and $\angle \mathrm{OPQ}=90^{\circ }$. We are given that $\mathrm{OQ}=12\mathrm{cm}$. By Pythagoras Theorem, we have ${\mathrm{PQ}}^2=O{Q}^2-{\mathrm{OP}}^2$ $=(12{)}^2-(5{)}^2=144-25=119$ $\Rightarrow \mathrm{PQ}=\sqrt{119}\mathrm{cm}$ Hence, the correct option is (D).
• Draw a circle and two lines parallel to a given line such that one is tangent and other a secant to the circle. Sol. We have the required figure as shown:

• Here, $\ell$ is the given line and a circle with centre 0 is drawn. The line n is drawn which is parallel to $\ell$ and tangent to the circle. Also, $m$ is drawn parallel to line $\ell$ and is a secant to the circle.

Exercise : 10.2

• From a point $Q$, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm . The radius of the circle is - (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm Sol. From figure,

• $r^2=(25{)}^2-(24{)}^2$ $=625-576=49$ r $=7\mathrm{cm}$ Hence, the correct option is (A).
• In fig., if TP and TQ are the two tangents to a circle with centre 0 so that $\angle \mathrm{POQ}=$ $110^{\circ }$, then $\angle \mathrm{PTQ}$ is equal to -

• (A) $60^{\circ }$ (B) $70^{\circ }$ (C) $80^{\circ }$ (D) $90^{\circ }$ Sol. TQ and TP are tangents to a circle with centre 0 and $\angle \mathrm{POQ}=110^{\circ }$ $\therefore \mathrm{OP}\perp \mathrm{PT}$ and $\mathrm{OQ}\perp \mathrm{QT}$ $\Rightarrow \angle \mathrm{OPT}=90^{\circ }$ and $\angle \mathrm{OQT}=90^{\circ }$ Now, in the quadrilateral TPOQ, we get $\therefore \angle \mathrm{PTQ}+90^{\circ }+110^{\circ }+90^{\circ }=360^{\circ }$ (Angle sum property of a quadrilateral) $\angle \mathrm{PTQ}+290^{\circ }=360^{\circ }$ $\angle \mathrm{PTQ}=360^{\circ }-290^{\circ }=70^{\circ }$ Hence, the correct option is (B)
• If tangents $PA$ and $PB$ from a point $P$ to a circle with centre 0 are inclined to each other at angle of $80^{\circ }$, then $\angle \mathrm{POA}$ is equal to (A) $50^{\circ }$ (B) $60^{\circ }$ (C) $70^{\circ }$ (D) $80^{\circ }$ Sol. In figure, $\Delta \mathrm{OAP}\cong \Delta \mathrm{OBP}$ (SSS congruence) $\Rightarrow \angle \mathrm{POA}=\angle \mathrm{POB}$ $=\frac{1}{2}\angle \mathrm{AOB}$ Also $\angle \mathrm{AOB}+\angle \mathrm{APB}=180^{\circ }$ $\Rightarrow \angle \mathrm{AOB}+80^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{AOB}=100^{\circ }$ Then from (i) and (ii) $\angle \mathrm{POA}=\frac{1}{2}\times 100=50^{\circ }$ Hence, the correct option is (A)
• Prove that the tangents drawn at the ends of a diameter of a circle are parallel. Sol. In the figure, PQ is diameter of the given circle and O is its centre. Let tangents $AB$ and $CD$ be drawn at the end points of the diameter PQ . Since, the tangents at a point to a circle is perpendicular to the radius through the point

• $\therefore \mathrm{PQ}\perp \mathrm{AB}$ $\Rightarrow \angle \mathrm{APQ}=90^{\circ }$ and $\mathrm{PQ}\perp \mathrm{CD}$ $\Rightarrow \angle \mathrm{PQD}=90^{\circ }$ $\Rightarrow \angle \mathrm{APQ}=\angle \mathrm{PQD}$ But they form a pair of alternate angles. $\therefore \mathrm{AB}\Vert \mathrm{CD}$. Hence, the two tangents are parallel.

• Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. Sol. Let us assume a circle with centre 0 and let $AB$ be the tangent intersecting the circle at point P . Also let us assume a point $X$ such that $XP$ is perpendicular to AB . We have to prove that XP passes through centre 0 . We know that Tangent of a circle is perpendicular to radius at point of contact. $\Rightarrow \mathrm{OP}\perp \mathrm{AB}$ (Therorem-1) So, $\angle \mathrm{OPB}=90^{\circ }$ We have already assumed that XP is perpendicular to AB . $\angle \mathrm{XPB}=90^{\circ }$ Now from equation (1) and (2) $\angle \mathrm{OPB}=\angle \mathrm{XPB}=90^{\circ }$ This condition is possible only if line XP passes through 0 . Since, XP passes through centre 0 . Therefore, it is proved that the the perpendicular at the point of contact of the tangent of a circle passes through the centre.
• The length of a tangent from a point $A$ at a distance 5 cm from the centre of the circle is 4 cm . Find the radius of the circle. Sol. The tangent to a circle is perpendicular to the radius through the point of contact. $\therefore \angle \mathrm{OTA}=90^{\circ }$ Now, in the right $△0TA$, we have : ${\mathrm{OA}}^2={\mathrm{OT}}^2+{\mathrm{AT}}^2$ [Pythagoras theorem]

• $\Rightarrow 5^2=0T^2+4^2$ $\Rightarrow {\mathrm{OT}}^2=5^2-4^2$ $\Rightarrow {\mathrm{OT}}^2=(5-4)(5+4)$ $\Rightarrow {\mathrm{OT}}^2=1\times 9=9=3^2$ $\Rightarrow \mathrm{OT}=3$ Thus, the radius of the circle is 3 cm .
• Two concentric circles are of radii 5 cm and 3 cm . Find the length of the chord of the larger circle which touches the smaller circle. Sol. In fig. the two concentric circles have their centre at 0 . The radius of the larger circle is 5 cm and that of the smaller circle is 3 cm .

• $AB$ is a chord of the larger circle and it touches the smaller circle at P . Join OA, OB and OP. Now, $\mathrm{OA}=\mathrm{OB}=5\mathrm{cm}$, $\mathrm{OP}=3\mathrm{cm}$ And $OP\perp AB$, i.e., $\angle \mathrm{OPA}=\angle \mathrm{OPB}=90^{\circ }$ $\Rightarrow \Delta \mathrm{OAP}\cong △\mathrm{OBP}$ (RHS congruence) $\Rightarrow \mathrm{AP}=\mathrm{BP}=\frac{1}{2}\mathrm{AB}$ or $\mathrm{AB}=2\mathrm{AP}$ By Pythagoras theorem, $O{A}^2=A{P}^2+{\mathrm{OP}}^2$ $\Rightarrow (5{)}^2={\mathrm{AP}}^2+(3{)}^2$ $\Rightarrow A{P}^2=25-9=16$ $\Rightarrow \mathrm{AP}=4\mathrm{cm}$ $\Rightarrow \mathrm{AB}=2\times 4\mathrm{cm}=8\mathrm{cm}$
• A quadrilateral ABCD is drawn to circumscribe a circle (see fig.). Prove that $AB+CD=AD+BC$.

• Sol. In fig., we observe that AP = AS $(\because$ AP and AS are tangents to the circle drawn from the point A) Similarly, BP = BQ $CR=CQ$ DR = DS Adding (i), (ii), (iii), (iv), we have $(\mathrm{AP}+\mathrm{BP})+(\mathrm{CR}+\mathrm{DR})=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ})$ $\Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}$ Note: This is a pitot theorem.
• In fig., $XY$ and $X^{\prime }{Y}^{\prime }$ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and $X^{\prime }{Y}^{\prime }$ at B. Prove that $\angle \mathrm{AOB}=90^{\circ }$

• Sol. In fig., Join OC and we have $△AOP$ and $△AOC$ for which $\Rightarrow \mathrm{AP}=\mathrm{AC}$ (Both tangents from A$)$ $\Rightarrow \mathrm{OP}=\mathrm{OC}$ (Each $=$ radius) $\Rightarrow 0A=0A$ (Common side) $\Rightarrow △\mathrm{AOP}\cong △\mathrm{AOC}$ (SSS congruence) $\Rightarrow \angle \mathrm{PAO}=\angle \mathrm{CAO}$ $\Rightarrow \angle \mathrm{PAC}=2\angle \mathrm{OAC}$ Similarly, $\angle \mathrm{QBC}=2\angle \mathrm{OBC}$ Adding (i) and (ii), $\angle \mathrm{PAC}+\angle \mathrm{QBC}=2(\angle \mathrm{OAC}+\angle \mathrm{OBC})$ $\Rightarrow 180^{\circ }=2(\angle \mathrm{OAC}+\angle \mathrm{OBC})$ $(\because$ sum of co-interior angles is $180^{\circ }$ ) $\Rightarrow \angle \mathrm{OAC}+\angle \mathrm{OBC}$ $=\frac{1}{2}\times 180^{\circ }=90^{\circ }$ Now, in $△AOB$ we have

$$\begin{gathered} \angle \mathrm{AOB}+\angle \mathrm{OAC}+\angle \mathrm{OBC}=180^{\circ } \\ \Rightarrow \angle \mathrm{AOB}+90^{\circ }=180^{\circ }(\mathrm{By}(\mathrm{iii})) \\ \Rightarrow \angle \mathrm{AOB}=90^{\circ } \end{gathered}$$

• Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. Sol. Let PA and PB be two tangents drawn from an external point P to a circle with centre 0 .

Now, in right $△\mathrm{OAP}$ and right $△\mathrm{OBP}$, we have, $\mathrm{PA}=\mathrm{PB}$ [Tangents to circle from an external point] $\mathrm{OA}=\mathrm{OB}$ [Radii of the same circle] OP = OP [Common] $\Delta \mathrm{OAP}\cong △\mathrm{OBP}$ [By SSS congruency] $\therefore \angle \mathrm{OPA}=\angle \mathrm{OPB}[$ [By C.P.C.T.] and $\angle \mathrm{AOP}=\angle \mathrm{BOP}$ $\Rightarrow \angle \mathrm{APB}=2\angle \mathrm{OPA}$ and $\angle \mathrm{AOB}=2\angle \mathrm{AOP}$ But $\angle \mathrm{AOP}=90^{\circ }-\angle \mathrm{OPA}$ $\Rightarrow 2\angle \mathrm{AOP}=180^{\circ }-2\angle \mathrm{OPA}$ $\Rightarrow \angle \mathrm{AOB}=180^{\circ }-\angle \mathrm{APB}$ $\Rightarrow \angle \mathrm{AOB}+\angle \mathrm{APB}=180^{\circ }$

• Prove that the parallelogram circumscribing a circle is a rhombus. Sol. Let ABCD be a parallelogram such that its sides touch a circle with centre 0 .

• $\begin{array}{rl}\Rightarrow & \mathrm{AP}=\mathrm{AS}[\text{ Tangents from an external }\\ & \text{ point are equal }]\end{array}$ $\Rightarrow \mathrm{BP}=\mathrm{BQ}$ [Tangents from an external point are equal] $\Rightarrow \mathrm{CR}=\mathrm{CQ}$ [Tangents from an external point are equal] $\Rightarrow \mathrm{DR}=\mathrm{DS}$ [Tangents from an external point are equal] Adding these equations, $\Rightarrow \mathrm{AP}+\mathrm{BP}+\mathrm{CR}+\mathrm{DR}=\mathrm{AS}+\mathrm{DS}+\mathrm{BQ}+\mathrm{CQ}$ $\Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}$ $\Rightarrow 2\mathrm{AB}=2\mathrm{BC}$ $\Rightarrow \mathrm{AB}=\mathrm{BC}$ and $\mathrm{AB}=\mathrm{DC}$ $\Rightarrow \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ $\Rightarrow \mathrm{ABCD}$ is a rhombus. Hence proved.
• A triangle $ABC$ is drawn to circumscribe a circle of radius 4 cm such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths 8 cm and 6 cm respectively (see fig.). Find the sides $AB$ and $AC$.

• Sol.

• In fig. $BD=8\mathrm{cm}$ and $DC=6\mathrm{cm}$ Then we have $\mathrm{BE}=8\mathrm{cm}(\because \mathrm{BE}=\mathrm{BD})$ and $CF=6\mathrm{cm}(\because CF=CD)$ Suppose AE = AF $=\mathrm{xcm}$ In $△ABC$, $\mathrm{a}=\mathrm{BC}=6\mathrm{cm}+8\mathrm{cm}=14\mathrm{cm}$ $\mathrm{b}=\mathrm{CA}=(\mathrm{x}+6)\mathrm{cm}$ $\mathrm{c}=\mathrm{AB}=(\mathrm{x}+8)\mathrm{cm}$ $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{14+(\mathrm{x}+6)+(\mathrm{x}+8)}{2}\mathrm{cm}$ $=\frac{2x+28}{2}\mathrm{cm}$ $=(x+14)\mathrm{cm}$ Area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{(x+14)\times x\times 8\times 6}$ $=\sqrt{48\mathrm{x}\times (\mathrm{x}+14)}{\mathrm{cm}}^2$ Also, area of $△ABC=$ area of $△OBC+$ area of $△OCA+$ area of $△OAB$ $=\frac{1}{2}\times 4\times a+\frac{1}{2}\times 4\times b+\frac{1}{2}\times 4\times c$ $=2(\mathrm{a}+\mathrm{b}+\mathrm{c})=2\times 2\mathrm{s}=4\mathrm{s}$ $=4(x+14){\mathrm{cm}}^2$ From (i) and (ii), $\sqrt{48\mathrm{x}\times (\mathrm{x}+14)}=4\times (\mathrm{x}+14)$ On squaring both sides $\Rightarrow 48\mathrm{x}\times (\mathrm{x}+14)=16\times (\mathrm{x}+14{)}^2$ $\Rightarrow 3\mathrm{x}=\mathrm{x}+14$ $\Rightarrow \mathrm{x}=7\mathrm{cm}$ Then $\mathrm{AB}=\mathrm{c}=(\mathrm{x}+8)\mathrm{cm}$ $=(7+8)\mathrm{cm}=15\mathrm{cm}$ and $\mathrm{AC}=\mathrm{b}=(\mathrm{x}+6)\mathrm{cm}$ $=(7+6)\mathrm{cm}=13\mathrm{cm}$
• Prove that opposite sides of $a$ quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Sol.

• Let $ABCD$ be a quadrilateral circumscribing a circle with centre 0 . Now join AO, BO, CO, DO. From the figure, $\angle \mathrm{DAO}=\angle \mathrm{BAO}$ [Since, AB and AD are tangents] Let $\angle \mathrm{DAO}=\angle \mathrm{BAO}=\angle 1$ Also $\angle \mathrm{ABO}=\angle \mathrm{CBO}$ [Since, BA and BC are tangents] Let $\angle \mathrm{ABO}=\angle \mathrm{CBO}=\angle 2$ Similarly, we take the same way for vertices $C$ and $D$. Recall that sum of the angles in quadrilateral, $\mathrm{ABCD}=360^{\circ }$ $=2(\angle 1+\angle 2+\angle 3+\angle 4)=360^{\circ }$ $=\angle 1+\angle 2+\angle 3+\angle 4=180^{\circ }$ In $△\mathrm{AOB},\angle \mathrm{BOA}=180^{\circ }-(\angle 1+\angle 2)$ In $△$ COD, $\angle \mathrm{COD}=180^{\circ }-(\angle 3+\angle 4)$ $\angle \mathrm{BOA}+\angle \mathrm{COD}=360^{\circ }-(\angle 1+\angle 2+\angle 3+\angle 4)$ $=360^{\circ }-180^{\circ }=180^{\circ }$ $\therefore \mathrm{AB}$ and CD subtend supplementary angles at 0 . $\because$ Sum of the angles at the centre is $360^{\circ }$. $\therefore \mathrm{AD}$ and BC subtend supplementary angles at 0 . Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.