SChand CLASS 9 Chapter 13 Circles Exercise 13(A)

 Exercise 13 A

Question 1

Sol: In the given circle radius $O A=13 \mathrm{~cm}$ and  ⊥$\mathrm{~OM}=12 \mathrm{~cm}$

(IMAGE TO BE ADDED)

Let Length of chord $A B=2 x$

$O m \perp A B$

So $m$ is midpoint of AB

$\text { So } A M=\frac{2 x}{2}=x$

Now in rignt angled $\triangle O A M$.

$O A^{2}=O M^{2}+A M^{2}$

$\Rightarrow(10)^{2}=(12)^{2}+x^{2} \Rightarrow 169=144+x^{2}$

$\Rightarrow x^{2}=169-144 \Rightarrow x^{2}-25=(5)^{2}$

so $x=5$

so Length of chord $A B=2 \times 5=10 \mathrm{~cm}$

Question 2

(IMAGE TO BE ADDED)

Sol: In the given circle

Radius $\mathrm{CA}=6 \mathrm{~cm}$ and $\perp \mathrm{cm}=3 \mathrm{~cm}$

Let length of $A B=x \mathrm{~cm}$

Then $A M C=\frac{1}{2} A B=\frac{x}{2} \mathrm{~cm}$

Now in right angled $\triangle C A M$

$C A^{2}=c m^{2}+A M^{2}$

$(6)^{2}=(3)^{2}+\left(\frac{x}{2}\right)^{2} \Rightarrow 36=9+\frac{x^{2}}{4}$

$\frac{x^{2}}{4}=36-9=27=(3 \sqrt{3})^{2}$

So $\frac{x}{2}=3 \sqrt{3} x=6 \sqrt{3}$

So Length of $A B=6 \sqrt{3} \mathrm{~cm}$

Question 3

Sol: In the figure,

C D is the diameter of the circle center $O A B$ is chord which  intersects $C D$ at $E$ and $E A=E B$ $=4 \mathrm{~cm}$ and $C E=3 \mathrm{~cm}$

Let r be the radius of the circle then 

OC = r and OE = (r-3)

Join OB 

In right $\triangle O B C$

$\begin{aligned}& O B^{2}=B E^{2}+O E^{2} \\\Rightarrow & r^{2}=(4)^{2}+(r-3)^{2} \Rightarrow r^{2}=16+r^{2}-6 r+9 \\\Rightarrow & r^{2}-r^{2}+6 r=16+9 \\\Rightarrow & 6 r=25 \Rightarrow r=\frac{25}{6}=4 \frac{1}{6}\end{aligned}$

Question 4

Sol: (i) When the chords are the same side of the center 

Radius of the circle =5cm

Length of AB = 8cm and CD = 6cm 

From O, draw a perpendicular on CD which Intersects AB at L and Meet CD at M 

So, M is the mid point of CD and L is the mid points of AB 

(IMAGE TO BE ADDED)

Join $O A$ and O C

 Now in right $\triangle \mathrm{OCM}$, 

$O C^{2}=c m^{2}+OM^{2}$ 

$\Rightarrow(5)^{2}=(3)^{2}+0 m^{2}$ 

$\Rightarrow 25=9+0 \mathrm{~m}^{2}$ 

$\Rightarrow \quad O M^{2}=25-9=16^{2}=4$

OM = 4cm

Similarly in right angled $\angle OAL$

$\begin{aligned} O A^{2} &=A L^{2}+O L^{2} \\ \Rightarrow(5)^{2} &=(4)^{2}+O L^{2} \\ 25 &=16+O L^{2} \Rightarrow O L^{2}=25-16=9=(3)^{2} \end{aligned}$

OL = 3cm

Now LM =OM -OL = 4.3 =1cm

So distance between the two chords =1cm

(ii) (IMAGE TO BE ADDED)

 Sol: When the chords are in the opposite sides of the center

Draw $O L \perp A B$ and $O M \perp C D$

In right $\angle OCM$

$\begin{aligned} & O C^{2}=c m^{2}+OM^{2} \\ \Rightarrow &(5)^{2}=(3)^{2}+OM^{2} \\ \Rightarrow & 25=9+OM^{2} \\ \Rightarrow & OM^{2}=25-9=16=(4)^{2} \\ & OM=4 c m \end{aligned}$

Similarly in right $\angle OLA$

$\begin{aligned} & O A^{2}=O L^{2}+A L^{2} \\ \Rightarrow &(5)^{2}=O L^{2}+(4)^{2} \\ \Rightarrow & 25=O L^{2}+16 \\ & O L^{2}=25-18=9=(3)^{2} \end{aligned}$

So OL = 3cm

So LM =OL + OM = 3+4=7cm

So distance between the two chords =7cm

Question 5

Sol: (IMAGE TO BE ADDED)

In right angled OAM, 

$A M=0.7 \mathrm{~cm}$ 

$O A=2.5 \mathrm{~cm}$ and $O A^{2}=\mathrm{AM}^{2}+\mathrm{OM}^{2}$

$\Rightarrow\left(2.51^{2}=(0.7)^{2}+OM^{2}\right.$

$\Rightarrow 6.25=0.49+OM^{2}$

$\Rightarrow OM^{2}=6.25-0.49=5.76=(2.4)^{2}$

so om $=2.4$

But $L M=3.9$

so $O \mathrm{~L}=\mathrm{LM}-O \mathrm{~M}=3.9-2.4=1.5 \mathrm{~cm}$

Now in  $\triangle O L C$

$\begin{aligned} & O C^{2}=C L^{2}+O L^{2} \\ \Rightarrow &(2.5)^{2}=C L^{2}+(1.5)^{2} \Rightarrow 6.25=C L^{2}+2.25 \\ C L^{2} &=6.25-2.25=4.00=(2)^{2} \end{aligned}$

So $\quad \mathrm{CL}=2$

So chord $C F=2 \times C L=2 \times 2=4 \mathrm{~cm}$

Question 6

Sol: (IMAGE TO BE ADDED)

Let r be the radius of the circle and chord AB = 18cm

and distance from the center O 

So OL = 2cm 

Let CD be another chord

 OM⊥ CD and OM = 6cm

Now in right $\triangle O A L$

$O A^{2}=A L^{2}+O L^{2}$

$=(9)^{2}+(2)^{2}=81+4=85$

Similarly in right $\angle OCM$

$O C^{2}=OM^{2}+CM^{2}$

$\Rightarrow O A^{2}=OM^{2}+c m^{2}$

$\Rightarrow 85=(6)^{2}+CM^{2}$

$\Rightarrow 85=36+CM^{2} \Rightarrow CM^{2}=85-36=49=(7)^{2}$

So $CM=7$

So $C D=2 \times CM=2 \times 7=14 \mathrm{~cm}$

Question 7

Sol: In the figure two circle are concentric with center O chord AB intersect the small circle at C and D 

OB = 17cm, OC = 10cm

$O L \perp A B$

Now in Right $\triangle$ OM $B$

$\begin{aligned}&O B^{2}=O M^{2}+M B^{2} \\&\Rightarrow(17)^{2}=\left(91^{2}+M B^{2}\right.\end{aligned}$

$\Rightarrow 289=81+M B^{2} \Rightarrow M B^{2}=289-81=208$

so $M B=\sqrt{208}=14.42$

or $\mathrm{AM}=14.42 \mathrm{~cm}$

Similarly in right $\triangle \mathrm{OCM}$

$\begin{aligned} & O C^{2}=OM^{2}+\mathrm{CM}^{2} \\ \Rightarrow &(10)^{2}=(9)^{2}+CM^{2} \Rightarrow 100=81+\mathrm{CM}^{2} \\ \Rightarrow & \mathrm{CM}^{2}=100-81=19 . \end{aligned}$

So CM = $\sqrt{19}=4.36 \mathrm{~cm}$

So AC = AM- CM = 1442 -4.36 = 10.06cm 

Question 8

Sol: Two equal circle with center O and O' intersect 

Each other at A and B

 (IMAGE TO BE ADDED)

OO' , AB , OA , OB , O'A and O'B are joined 

AB = 10cm 

OO' = 6cm

Let OO' Bisects AB at M i.e AM =MB = $\frac{10}{2}=5 \mathrm{~cm}$

and $OM=MO'=\frac{6}{2}=3 \mathrm{~m}$

Now right $\triangle O AM$

$O N^{2}=OM^{2}+NM^{2}$

$=(3)^{2}+(5)^{2}=9+25=34$

$O A=\sqrt{34}=5.83 \mathrm{~cm}$

Hence radius of each circle = 5.84cm

Question 9

Sol: In circle with center O. OA is its radius 

AB is chord OM ⊥ AB Which is produced to meet the circle at C

$A B=8 \mathrm{~cm}, \mathrm{~cm}=1 \mathrm{~cm}O A=X$ 

$O C=O A=X$

So OM= OC- CM =(X-1)

Now in right $\angle OAM$

$O A^{2}=A M^{2}+O M^{2}$

$X^{2}=(4)^{2}+(X-1)^{2}$

$\Rightarrow x^{2}=16+x^{2}-2 x+1$

$\Rightarrow X^{2}-X^{2}+2 x=17$

$\Rightarrow 2 x=17 \Rightarrow x=\frac{17}{2}=8.5$

Hence x = 8.5cm

Question 10

 (IMAGE TO BE ADDED)

Sol: In the circle  with center O. AB is chord CD is perpendicular bisector of AB

$A B=2 \mathrm{~cm}, C D=4 \mathrm{~cm}$

Join OA, Let r be the radius of the circle 

OA = OD =r 

OC =CD -OD =4-r

$A C=\frac{1}{2}, A B=\frac{1}{2} \times 2=1 \mathrm{~cm}$

Now in right $\triangle O A C$

 $O A^{2}=O C^{2}+A C^{2}$

$\Rightarrow r^{2}=(4-r)+(1)^{2} \Rightarrow r^{2}=16+r^{2}-8 r+1$

$\Rightarrow r^{2}-r^{2}+8 r=17 \Rightarrow 8 r=17$

$\Rightarrow r=\frac{17}{8}=2 \frac{1}{8}$

So radius of the circle = $2 \frac{1}{8} \mathrm{~cm}$

Question 11

Sol: In the circle $A B$ is chord $O D \perp A B B O C$ is the diameter of the circle

$A C$ is joined 

To prove: $C A=2OD$

 proof : if $OD \perp A B$

D mid point of AB and O is mid point of BC 

So In $\triangle A B C, D O \| A C$

So In $\triangle A B C-\triangle D B O$

So $\begin{aligned} & \frac{A B}{B D}=\frac{B C}{B O}=\frac{A C}{O D} \\ \Rightarrow & \frac{A B}{\frac{1}{2} A B}=\frac{C A}{O D} \Rightarrow \frac{2}{1}=\frac{C A}{O D} \Rightarrow C A=2 O D \end{aligned}$

Hence proved 

Question 12

 (IMAGE TO BE ADDED)

Sol: In the figure 

OMNP is a square 

A circle with center O is drawn which intersect the square at X and Y 

Join OX and OY

Prove: 

(i) $\triangle OXM \cong \triangle OYP$

(ii) NX = NY 

Proof : In right $\triangle OXM$ and \triangle OYP

Hyp  $OX=O Y$ 

side $O M=O P$

$\triangle OXM \cong \triangle OYP$

MX = PY 

But MN = PN 

so MN -MX =PN -PY

=NX =NY Hence proved

Question 13

Sol:(i) Take three points $A_{1} B$ and $C$ on the circle

(ii) Join $A B$ and $A C$

(iii) Draw the perpendicular bisectors of $A B$ and $A C$. which intersect each other at O.

Then O is the center of the circle (We know that perpendicular bisector of chords of a circle passes through the center of the circle )

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Question 14

Sol: steps of construction:

(i) Take three points $A, B$ and $C$.

(ii) Join $A B$ and $B C$

(iii) Draw the perpendicular bisector of $A B$ and $B C$ intersecting each other at 0 .

(lv) with center $O$ and radius $O A$ or $O B$ or $O C$, draw an arc $A B C$.

This  is the required arc 

If O lies on the perpendicular bisector of AB. 

So OA = OB 

again O Lies on the perpendicular bisector of BC 

So OB = OC 

So OA =OB =OC

 (IMAGE TO BE ADDED)

Question 15

Ans: False : As through three points which are not in a straight line, one and only once circle can be drawn. 

Question 16

Sol: $A B$ and $C D$ are two paraltel chords as the circle and a line  the perpendicular bisector of AB

i.e. $\quad A L=\angle B$ and $\angle A L M=90^{\circ}$

(IMAGE TO BE ADDED)

To prove: Line $L$ is also perpendicular bisector of $C D$

 proof: if line $L$ is perpendicular bisector of chord $A B$ 

so it will pass through the center of the circle 

if $A B \| C D$ and $L$ is perpendicular to $A B$

So $L$ is also perpendicular to $C D$

if $O m \perp C D$

so M is the mid point of C D

Hence proved

Question 17

Sol: Two circle with center O and O' intersect each other at P. 

APB||OO' is drawn 

To prove : AB = 2OO' 

Construction: From O, Draw OM ⊥ AP 

So MN = OO'

Now if OM ⊥  AP 

So M is mid point of AP 

AM = MP = MP $=\frac{1}{2} A P$

similarly $P N=N B \Rightarrow P N=\frac{1}{2} P B$

Adding we get

$m P+P N=\frac{1}{2} A P+\frac{1}{2} P B$

$\Rightarrow M N=\frac{1}{2}(A P+P B)=\frac{1}{2} A B$

$\begin{aligned}&\Rightarrow O O^{\prime}=\frac{1}{2} A B \\&\Rightarrow 200^{\prime}=A B \\&\text { or } A B=2OO^{\prime}\end{aligned}$

Question 18

Sol: In a circle with center O, two equal chords AB and CD intersect each other at P

To prove: AP = PD and BP = CP 

Construction: Draw OM⊥ AB and ON ⊥CD

Proof: In right $\triangle OMP$ and $\triangle O N P$

Hyp OP =OP 

OM = ON 

(Equal chords are equidistant from the center)

so $\triangle O M P \cong \triangle O N P$

so $P M=P N$

if $O M \perp A B$ and $O N \perp C D$

So M and N are the mid points of AB and CD respectively 

Or AM =MP and CN = NP 

So chords AB =CD 

So AB - PB = CD -PC 

=AP =PD 

Similarly BM - PM = CN -PN 

BP=CP Hence proved

Question 19

SoL: In a circle with centre 0. chord $A B=C D$

$O E \perp C D$ at $A$ and $O F \perp A B$ at $G$

To prove: $E H=h F$

Proof: if $O E \perp C D$ and $O F \perp A B$

So G and H are mid points of AB and CD respectively 

If and OG=OH 

(Equal chords are equidistant from the center )

But of $=O E$

So $O F-O G=O E-O H$

$\Rightarrow G F=E H$

So $E H=G F$

Hence proved 

Question 20

Sol: In circle with conter C

$C B$ bisects $\angle D B E$

$C D \perp P Q$ and $C E \perp R S$ where $P Q$ and $R s$ are the chord which intersect each other at B.

To prove: $P Q=R S$ 

proof: In $\triangle E B C$ and $\triangle D B C$.

$\angle E=\angle D$

$\angle E B C=\angle D B C$

$B C=B C$

So $\triangle E B C \cong \triangle D B C$

So and $C E=C D$

But these are the perpendicular distance from the center and equal chords are equidistant from the center 

So RS $=P Q$

or $P Q=R S$ Hence proved

Question 21

Sol: In circle with center O. 

Two equal chords AB and CD intersect each other inside the circle 

$O M \perp A B$ and $O N \perp C D$ 

M N are joined

To prove: $\angle O M N=\angle O N M$ 

proof: is chord $A B=C D$

So OM=ON

(Equal chords are equidistant from the center)

In $\triangle OMN$

If OM = ON 

So $\angle OMN$= $\angle ONM$ Hence proved