A circle is a simple closed geometrical shape equidistant from a central point. It is an important shape in the field of geometry. Every circle has its center.

The straight line from the center to the circumference of a circle is called the radius of the circle. The length of the line through the center that touches two points on the edge of the circle is called a diameter. The total distance around the circle is called as Circumference.

The area of the circle can be calculated by applying the formula:

A = π r2

Where: A is the Area, r is the radius and the value of π is 3.14.

With the help of our downloadable NCERT Solutions for Class 9 Maths Chapter 10 circles, students can try solving textbook problems by themselves and also correct themselves with the solutions provided with the right steps. Some of the chapters included in the NCERT Solutions For Class 9 Maths Chapter 10 Circles are Introduction to circles, definitions of circle related concepts, arc, radius, diameter, circumference, chord, sector, secant, segment subtended angle. Here students can learn a lot more related to the topic Circles along with the various examples and problems. For more information about the NCERT Solutions for Class 9 Maths Chapter 10 circles, students can check the below-mentioned pdf files.

**Short Answer Questions**

*1. Out of the two concentric circle ,the radius of the outer circle is 5cm and the chord FC is of length 8cm is a tangent to the inner circle .Find the radius of the inner circle.*

Sol. Let the chord FC of the larger circle touch the smaller circle at the point L.

Since FC is tangent at the point L to the smaller circle with the centre O.

∴ OL\(\perp\)FC

Since AC is chord of the bigger circle and OL\(\perp\)FC.

∴ OL bisects FC

∴ FC=2FL

\(\Rightarrow\) 8=2FL

\(\Rightarrow\) FL=4cm

Now, consider right-angled \(\Delta\)FLO,we obtain

\(OL^{2}\)= \(FO^{2}\)- \(FL^{2}\)

= \(5^{2}\)-\(4^{2}\)

=25-16

=9

OL =\(\sqrt{9}\)=3

Hence, the radius of the smaller or inner circle is 3cm

*2. Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.*

Sol. Let \(l_{1}\) and \(l_{2}\),two intersecting lines, intersect at A, be the tangents from an external point A to a circle with centre X, at B and F respectively.

Join XB and XF

Now, in \(\Delta\)ABX and \(\Delta\)ARX, we have

AX=AX (common)

XB=XF (radii of same circle)

AB=AF

(tangents from an external point)

∴ \(\Delta\)ABX=\(\Delta\)AFX

(by SSS congruence rule)

\(\Rightarrow\) \(\angle\)XAB=\(\angle\)XAF

\(\Rightarrow\) X lies on the bisector of the lines \(l_{1}\) and \(l_{2}\).

*3.In the given figure,LM and NP are common tangents to two circles of unequal radii.Prove that LM=NP.*

Sol. Let Chords LM and NP meet at the point R

Since RL and RN are tangents from an external point R to two Circles with centres O and \({O}'\).

∴ RL=RN …..(i)

And RM=RP …..(ii)

Subtracting (ii) from (i),we have

RL-RM=RN-RP

∴ LM=NP

*4.In the given figure, common tangents PQ and RS to two circles intersect at T. Prove that PQ=RS.*

Sol. Clearly, TP and TR are two tangents from an external point T to the circle with centre O.

∴ TP=TR ……(i)

Also, TQ and TS are two tangents from an external point T to the circle with centre \({O}'\).

∴ TQ=TS …..(ii)

Adding (i) and (ii),we obtain

TP+TQ=TR+TS

\(\Rightarrow\) PQ=RS

*5.A chord XY of a circle is parallel to the tangent drawn at a point Z of the circle. Prove that Z bisects the arc XZY.*

Sol. Since XY is parallel to the tangent drawn at the point Z and radius OZ is perpendicular to the tangent.

∴ OR\(\perp\)XY

∴ OL bisects the chord XY.

∴ XL-LY

∴ arc XZ=arc ZY

i.e., Z bisects arc XZY

__Long answer questions __

*1.If from an external point*__A__*of a circle with centre**X**,**two tangents AC** and **AD are drawn such that **\(\angle\)DAC=120°** , prove that AC+AD=AX** i.e.,**AX=2AC.*

*Sol*. Join XC and XD.

since, XCAC

∴ [tangent to any circle is perpendicular to its radius at point of contact]

∴ \(\angle\)XCA=90°

In \(\Delta XCB\) and \(\Delta XDB\),we have

CA=DA

[tangents from an external point]

XA=XA [common side]

XC=XD [radii of a circle]

∴ By SSS congruency , we have

\(\Delta XCA\)\(\cong\)\(\Delta XDA\)

\(\Rightarrow \cong\) \(\angle\)XAC=\(\angle\)XAD

=\(\frac{1}{2}\)\(\angle\)CAD=\(\frac{1}{2}\)\(\times\)\(120^{\circ}\)=\(60^{\circ}\)

In \(\Delta XCA\) ,\(\angle\)XCA=90°

∴ \(\frac{AC}{AX}=\cos 60^{\circ}\)

\(\Rightarrow \frac{BC}{BX}=\frac{1}{2}\)

\(\Rightarrow\) AX=2AC

Also, AX=AC + AC

\(\Rightarrow\) AX=AC + AD [Since, AC=AD]

*2.If x,y,z are the sides of a right triangle where c is hypotenuse,prove that the radius r of the circle which touches the sides of the triangle is given by*

* *

* r = \(\frac{x+y-z}{2}\)*

Sol. Let the circle touches the sides YZ,ZX,XY of the right triangle XYZ at D,E,F respectively,where YZ=x,ZX=y and XY=z.Then XE = XF and YD = YF.

Also, ZE=ZD=r

I.e., y-r=XF,

x-r=yF

Or XY=z=XF+YF

\(\Rightarrow\) Z=y-r+x-r

\(\Rightarrow\) r = \(\frac{x+y-z}{2}\)

*3.In the given figure,from an external poiny P,a tangent PT and a line segment PXY is drawn to a circle with the centre O.ON is perpendicular on the chord XY. Prove that:*

* (i)PX.PY=**\(PY^{2}\)-\( XN^{2}\)*

* (ii) \(PN^{2}- XN^{2}\)= \(OP^{2}- OT^{2}\)*

* (iii) \(PX.PY= PT^{2}\)*

Sol. (i) PX.PY=(PN-XN)(PN+YN)

=(PN-XN)(PN+XN)

since, XN=YN,as ON\(\perp\)Chord XY]

= \(PN^{2}- XN^{2}\)

(ii) \(PN^{2}- XN^{2}\)=(\(OP^{2}\)\(ON^{2}\))- \(XN^{2}\)

Since,in rt. \(\triangle\)ONP \(OP^{2}\)= \(ON^{2}\)+ \(PN^{2}\)]

=\(OP^{2}\)- \(ON^{2}\)- \(XN^{2}\)

=\(OP^{2}\)-(\(ON^{2}\))+ \(XN^{2}\))

=\(OP^{2}\)- \(OX^{2}\)

=\(OP^{2}\)- \(OT^{2}\)

Since,OX=OT=r]

(iii)From (i) and(ii),we have

PX.PY= \(OP^{2}\)- \(OT^{2}\)

\(PX.PY= PT^{2}\)

∴ In rt. \(\triangle\)OTP; \(OP^{2}\)= \(OT^{2}\)+ \(PT^{2}\)

*4.XY is a diameter and XZ is a chord with centre O such that \(\angle\)YXZ=**\(30^{\circ}\).The tangent at Z intersects XY extended at a point D. Prove that YZ=YD.*

*Sol. Here, XOY is a diameter of the circle, such that \(\angle\)YXZ=\(30^{\circ}\) and ZD be the tangent at Z.*

\(\angle\)XZY=\(90^{\circ}\)\(\angle\)in a semi-circle]

Also, \(\angle\)ZYD=\(\angle\)YXZ+\(\angle\)XZY

[ext. \(\angle\)of a \(\triangle\) is equal to sum of interior opp.angles]

=\(30^{\circ}\)+ \(90^{\circ}\)

=\(120^{\circ}\)

\(\angle\)YZD=\(\angle\)YXZ=\(30^{\circ}\)

In \(\triangle\),by angles sum property,we have

\(\angle\)YDZ+\(\angle\)ZYD+\(\angle\)YZD=\(180^{\circ}\)

\(\angle\)YDZ+\(120^{\circ}\)+\(30^{\circ}\)=\(180^{\circ}\)

\(\angle\)YDZ=\(180^{\circ}\)-\(120^{\circ}\)-\(30^{\circ}\)

=\(30^{\circ}\)

\Rightarrow \(\angle\)YDZ=\(\angle\)YDZ

=\(30^{\circ}\)

Hence, YD=YZ

[sides opp.to equal angles]

* *

*5.Two circles with centres O and \({O}'\) of radii 3 cm and 4cm respectively intersect at two points P and Q such that \({O}'\)P are tangents to the two circles.Find the length of the common chord PQ.*

Sol. Clearly, \(\angle\)OP\({O}'\)= \(90^{\circ}\)

O\({O}'\)= \({\sqrt{3^{2}+4^{2}}}'\)

=\(\sqrt{9+16}=\sqrt{25}\)

=5cm

Let RO=x

⇒ R\({O}'\)=5-x

Now, \(PR^{2}\)= \(PO^{2}\)- \(RO^{2}\)

=9-\(x^{2}\) ……..(1)

Also, \(PR^{2}\)= \(PO^{2}\)- \(RO^{2}\)

=16-\(\left ( 5-x \right )^{_{2}}\)……….(2)

From (1) and (2),we have

9-\(x^{2}\)=16-(25+\(x^{2}\)-10x)

9-\(x^{2}\)=16-25+\(x^{2}\)+10x)

10x=18

X=1.8cm

From (1),we have

\(PR^{2}\)=9-\(1.8^{2}\)

=9-3.24

=5.76

PR=\(\sqrt{5.76}\)

=2.4cm

Hence, the required length of the common chord is 2 x PR i.e., 2×2.4 i.e., 4.8cm.

__ __

__Exemplar problems__

*1. If V1 , V2 (V2 > V1) be the diameters of two concentric circles and Z be the length of a chord of a circle which is tangent to the other circle, prove that : \({V_{2}}^{2}=Z^{2}+{V_{1}}^{2}\).*

Solution:-

Let XY be a chord of a circle which touches the other circle at Z. Then \(\Delta\)OZY is a right triangle.

∴ By Pythagoras Theorem,

\(OZ^{2}\) + \(ZY^{2}\)= \(OY^{2}\)

i.e., \(\left ( \frac{V1}{2} \right )^{2}\)+ \(\left ( \frac{Z}{2} \right )^{2}\)= \(\left ( \frac{V2}{2} \right )^{2}\)

\(\Rightarrow\) \(\frac{{V_{1}}^{2}}{4}\)+ \(\frac{Z^{2}}{4}\)= \(\frac{{V_{2}}^{2}}{4}\)

\(\Rightarrow\) \({V_{2}}^{2}+Z^{2}={V_{1}}^{2}\)

*2. Two tangents AB and AC are drawn from an external point to a circle with centre 0. Prove that BOCA is a cyclic quadrilateral.*

Sol. We know that, tangents to a circle is perpendicular to its radius at the point of contact.

∴ OC\(\perp\)AC and OB\(\perp\)AB

\(\angle\)OCA=\(\angle\)OBA=\(90^{\circ}\)

\(\angle\)OCA+\(\angle\)OBA=\(90^{\circ}\)+ \(90^{\circ}\)

=\(180^{\circ}\)

Sum of opposite angles of quadrilateral BOCA is \(180^{\circ}\)

Hence, BOCA is a cyclic quadrilateral.

As mentioned earlier, circles are the most important and essential topic as it explains in detail about the circles, circumference, and radius its formulas and its constructions steps. BYJU’S provide free NCERT solutions for all the classes from the lower grade to the higher level of education. Our experts have designed these solutions to promote a better understanding of some basic concepts included in this chapter by illustrating solved questions, shortcut tips, important formulas and more. Here, in this NCERT solutions, students can gain more information about this topic as there are more solved example problems for students benefits. Interested students can either download the free PDF files from our website or can practice various questions related to this chapter by visiting our website at BYJU’S along with the interesting YouTube videos on Geometry and Circles.