[3 marks sum]

Question 13 Marks

Suppose you are given a circle. Describe a method by which you can find the center of this circle.

Answer

To draw the center of a given circle :
$1.$ Draw the circle.
$2.$ Take any two different chords $AB$ and $CD$ of this circle and draw perpendicular bisector of these chords.
$3.$ let these perpendicular bisectors meet at point $O.$
So, $O$ will be the center of the given circle.

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Question 23 Marks

Draw two circles of different radii. How many points these circles can have in common? What is the maximum number of common points?

Answer

So, the circle can have $0, 1$ or $2$ points in common.
The maximum number of common points is $2.$

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Question 33 Marks

In the following figure, the line $\text{ABCD}$ is perpendicular to $PQ;$ where $P$ and $Q$ are the centers of the circles.
Show that:$(i) AB = CD ;(ii) AC = BD.$

Answer

In the circle with center $Q, QO \perp AD$
$\therefore OA = OD ...(i) ...[$perpendicular drawn the center of a circle to a chord bisects it$]$
In circle with center $P, PO \perp BC$
$\therefore OB = OC ....(ii) ....[$perpendicular drawn the center of a circle to a chord bisects it$]$
$(i) - (ii)$ gives,
$AB = CD ...(iii)$
$(ii) $Adding $BC$ to both sides of equation $(iii)$
$AB + BC + CD + BC$
$\Rightarrow AC = BD$

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Question 43 Marks

The figure shows two concentric circles and $AD$ is a chord of a larger circle.
Prove that: $AB = CD.$

Answer

Drop $OP \perp AD$
$\therefore OP$ bisects $AD. ....($ Perpendicular drawn from the centre of a circle to a chord bisects it. $)$
$\Rightarrow AP = PD .....(i)$
Now, $BC$ is a chord for the inner circle and $OP \perp BC.$
$\therefore OP$ bisects $BC ....($ Perpendicular drawn from the centre of a circle to a chord bisects it. $)$
$\Rightarrow BP = PC .....(ii)$
Subtracting $(ii) $from $(i),$
$AP - BP = PD - PC$
$\Rightarrow AB = CD.$

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Question 53 Marks

The figure given below shows a circle with center $O$ in which diameter $AB$ bisects the chord $CD$ at point $E$. If $CE = ED = 8 \ cm$ and $EB = 4 \ cm,$
find the radius of the circle.

Answer

Let the radius of the circle be $r \ cm.$
$\therefore OE = OB - EB = r - 4$
Join $OC.$
In right $\triangle OEC,$
$OC^2 = OE^2 + CE^2$
$\Rightarrow r^2 = ( r - 4 )^2 + (8)^2$
$\Rightarrow r^2 = r^2 - 8r + 16 + 64$
$\Rightarrow 8r = 80$
$\therefore r = 10 \ cm$
Hence, radius of the circle is $10 \ cm.$

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Question 63 Marks

A chord of length $6 \ cm$ is drawn in a circle of radius $5 \ cm.$Calculate its distance from the center of the circle.

Answer

Let $AB$ be the chord and $O$ be the center of the circle.
Let $OC$ be the perpendicular drawn from $O$ to $AB.$

We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.
$\therefore AC = CB = 3 \ cm$
In $\triangle OCA,$
$OA^2 = OC^2 + AC^2 ...($ By Pythagoras theorem$ )$
$\Rightarrow OC^2 = (5)^2 - (3)^2$
$\Rightarrow OC = 16$
$\Rightarrow OC = 4 \ cm$

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Question 73 Marks

In the given figure, $O$ is the center of the circle. $AB$ and $CD$ are two chords of the circle. $OM$ is perpendicular to $AB$ and $ON$ is perpendicular to $CD . A B=24 \ cm, OM=5 \ cm, ON=12 \ cm$,

Find the $:(i)$ the radius of the circle$,(ii)$ length of chord $CD.$

Answer

$(i) AB$ is the chord of the circle and $OM$ is perpendicular to $AB.$
So, $AM = MB = 12 \ cm ....($ Since $\perp $ bisects the chord $)$
In right $\triangle OMA,$
$OA^2 = OM^2 + AM^2$
$\Rightarrow OA^2 = 5^2 + 12^2$
$\Rightarrow OA = 13 \ cm$
So, radius of the circle is $13 \ cm.$
$(ii)$ So, $OA = OC = 13 \ cm ....($ radii of the same circle $)$
In right $\triangle ONC,$
$NC^2 = OC^2 - ON^2$
$\Rightarrow NC^2 = 13^2 - 12^2$
$\Rightarrow NC = 5 \ cm$
So, $CD = 2NC$
$= 10 \ cm.$

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MATHEMATICS [3 marks sum]

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