Question 1: A chord of length $6 \ cm$ is drawn in a circle of radius $5 \ cm$. Calculate its distance from the center of the circle.

Let the distance from the center $= x$

Therefore $x = \sqrt{5^2-3^2}= \sqrt{25-9} = 4 \ cm$

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Question 2: A chord of length $8 \ cm$ is drawn  at a distance of $3 \ cm$ from the center of a circle. Calculate the radius of the circle.

Let the radius $= r \ cm$

Therefore $r = \sqrt{4^2+3^2} = 5 \ cm$

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Question 3: The radius of a circle is $17 \ cm$ and the length of perpendicular drawn from the center to a chord is $8 \ cm$. Calculate the length of the chord.

Let the length of the chord $= 2x$

Therefore $x = \sqrt{17^2-8^2} =\sqrt{289-64} = 15 \ cm$

Therefore the length of the chord $= 2x = 30 \ cm$

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Question 4: A chord of length 24 cm is at a distance of 5 cm from the center of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the center.

Let the radius $= r \ cm$

Therefore $r = \sqrt{12^2+5^2} = 13 \ cm$

Let the length of the chord $= 2x$

Therefore $x = \sqrt{13^2-12^2} =\sqrt{169-144} = 5 \ cm$

Therefore the length of the chord $= 2x = 10 \ cm$

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Question 5: In the following figure, $AD$ is a straight line $OP \perp AD$ and $O$ is the center of both the circles. If $OA = 34 \ cm, OB = 20 \ cm$ and $OP = 16 \ cm$, find the length of $AB$.

$BP = \sqrt{20^2-16^2}=\sqrt{400-256}= 12 \ cm$

$AP = \sqrt{34^2-16^2}=\sqrt{1156-256}= 30 \ cm$

Therefore $AB = (AP-BP) = 30 - 12 = 18 \ cm$

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Question 6: $O$ is the center of the circle of radius $10 \ cm$. $P$ is any point in the circle such that $OP=6 \ cm$. $A$ is the point travelling along the circumference, $x$ is the distance from $A to P$. What are the least and the greatest values of $x \ in \ cm$. What is the position of the points $O, P \ and \ A$ at these values . [1992]

When $P$ is on $OA$, the least distance of $x = 4 \ cm$

When $P$ is on extended $OA$, the greatest distance of $x = 16 \ cm$

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Question 7: In a circle of radius $17 \ cm$, two parallel chords of length $30 \ cm$ and $16 \ cm$ are drawn. Find the distance between the chords, if both the chord are:

(i) on the opposite side of the center

(ii) on the same side of the center

(i) When the chords are on the opposite side of the triangles

Distance of the larger chord from the center $x_1 = \sqrt{17^2-15^2}= 8 \ cm$

Distance of the smaller chord from the center $x_2 = \sqrt{17^2-8^2}= 15 \ cm$

Therefore the distance between the chords $= x_1 + x_2 = 23 \ cm$

(ii)  When the chords are on the same side of the triangles

Distance of the larger chord from the center $x_1 = \sqrt{17^2-15^2}= 8 \ cm$

Distance of the smaller chord from the center $x_2 = \sqrt{17^2-8^2}= 15 \ cm$

Therefore the distance between the chords $= x_2 - x_1 = 7 \ cm$

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Question 8: Two parallel chords are drawn in a circle of diameter $30 \ cm$. The length of one chord is $24 \ cm$. and the distance between the two chords is $21 \ cm$. Find the length of another chord.

Let the distance of the $24 \ cm$ chord from the center $= x \ cm$

Therefore $x = \sqrt{15^2-12^2}= 9 \ cm$

Therefore the distance of the other chord from the center $= 21 - 9 = 12 \ cm$

Therefore the length of the other chord $= 2 \times \sqrt{15^2-12^2} = 18 \ cm$

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Question 9: A chord $CD$ of a circle whose center is $O$, is bisected at $P$ by a diameter $AB$. Given  $OA = OB = 15 \ cm$ and $OP = 9 \ cm$, calculate the lengths of  (i) $CD$ (ii) $AD$ and (iii) $CB$

$CP = \sqrt{15^2-9^2} = 12 \ cm$

$CD = 2 \times 12 = 24 \ cm$

$AD = \sqrt{24^2+12^2}=\sqrt{720} = 26.83 \ cm$

$CB = \sqrt{12^2+(15-9)^2} = \sqrt{144+36} = 13.42 \ cm$

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Question 10: 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.

Let $OE = x \ cm$

Therefore

$x^2+8^2 = (x+4)^2$

$x^2 + 64 = x^2 + 8x + 16$

$8x = 48 \Rightarrow x = 6 \ cm$

Therefore Radius $= 6 + 4 = 10 \ cm$

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Question 11: The figure shows two concentric circles and $AD$ is the chord of larger circle. Prove that $AB = CD$.

If you drop a perpendicular from $O$ to the chord, it would bisect the chord (Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.). Let us say that it intersects $BC$ and $AD$ at $E$.

Therefore $BE = CE$ and $AE = DE$

If you subtract these two, we get

$AE - BE = DE - CE \Rightarrow AB = CD$

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Question 12: A straight line is drawn cutting two equal circles and passing through the midpoint $M$ of the line joining their centers $O \ and \ O'$.  Prove that the chords $AB \ and \ CD$, which are intercepted by two circles are equal.

First draw perpendiculars $OP$ and $O'P'$

Now consider $\triangle OPM$ and $\triangle O'P'M$

$\angle OMP = \angle P'MO'$ (opposite angles)

$\angle OPM = \angle O'P'M = 90^o$

$OM = O'M (M$ is the mid point of $OO'$ – Given)

Therefore  $\triangle OPM \cong \triangle O'P'M$

Therefore $OP = O'P'$

Therefore $AB = CD$ (Theorem 8 (Converse of Theorem 7) : Chords of circle which are equidistant from the center of the circle are equal in length.)

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Question 13: $M \ and \ N$ are mid points of two equal chords $AB \ and \ CD$ respectively of a circle with center $O$. Prove that:

(i) $\angle BMN = \angle DNM$

(ii) $\angle AMN = \angle CNM$

First drop perpendiculars $OM$ and $ON$ on $AB$ and $CD$ respectively.

$OM$ will bisect $AB$ and $ON$ will bisect $CD$. (Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.)

Therefore $BM = \frac{1}{2} AC and DN = \frac{1}{2}CD$

Given $AB = CD$. Therefore $BM = DN$

Also $OM^2 = OB^2-MB^2 = OD^2-DN^2 = ON^2$

$\Rightarrow OM = ON$

Therefore in $\triangle OMN, \angle OMN = \angle ONM$ (since angles opposite equal sides of a triangle are equal)

We know $\angle OMB = \angle ONC$

(i) Therefore $\angle OMB - \angle OMN = \angle ONC - \angle ONM$

$\Rightarrow \angle BOM = \angle DNM$

(ii) We know $\angle AMO = \angle CNO = 90^o$

Therefore $\angle AMO + \angle OMN = \angle CNO + \angle ONM$

$\Rightarrow \angle AMN = \angle CNM$

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Question 14: In the following figure; $P$ and $Q$ are the points of intersection of two circles with centers $O$  and $O'$. If straight lines $APB$ and $CQD$ are parallel to $OO'$; prove that

(i) $OO'=\frac{1}{2} AB$

(ii) $AB = CD$

$MP = \frac{1}{2} AP \ and \ NP = \frac{1}{2} PB$

$M'Q = \frac{1}{2} CQ \ and \ N'Q = \frac{1}{2} DQ$

$OO'=MN = MP+NP = \frac{1}{2} (AP+BP) = \frac{1}{2} AB$ … … … … (i)

$OO'=M'N' = M'Q+N'Q = \frac{1}{2} (CQ+DQ) = \frac{1}{2} CD$ … … … … (ii)

Therefore from (i) and (ii) we get $AB = CD$. Hence proved.

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Question 15: Two equal chords $AB$ and $CD$ of a circle with center $O$, intersect each other at point $P$ inside the circle. Prove that:

(i)  $AP = CP$

(ii) $BP = DP$

Draw $ON \perp CD$ and $OM \perp AB$

(Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.)

$MB = AM \Rightarrow AM = \frac{1}{2} BA$

$CN = ND \Rightarrow CN = \frac{1}{2} CD$

$\Rightarrow MB = CN = ND$

In $\triangle OMB \ \ OM^2 = OB^2 -MB^2$

In $\triangle OND \ \ ON^2 = OD^2 -ND^2$

$OM^2 = OB^2-ND^2$

$ON^2 = OD^2-ND^2$

$OB = OD$ (radius of the circle)

$\Rightarrow OM = OM$

Consider $\triangle OPM and \triangle OPN$

$OM = ON$

$\angle OMP = \angle ONP = 90^o$

$OP$ is common

Therefore  $\triangle OPM \cong \triangle OPN$

$\Rightarrow PM = PN$

$PM+MB = PN+ND \Rightarrow PB = DP$

$AB = CD$ (Given)

Since $BP = DP$

Therefore $AB - BP = CD - DB \Rightarrow AP = CP$

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Question 16: In the following figure, $OABC$ is a square. A circle is drawn with $O$ as the center which meets $OC$ at $P$ and $OA$ at $Q$. Prove that:

(i) $\triangle OPA \cong \triangle OQC$

(ii) $\triangle BPC \cong \triangle BQA$

(i) In $\triangle OPA \ and \ \triangle OQC$

$OP=OQ$ (radius)

$\angle AOP = \angle QOC = 90^o$

$OA = OC$ (side of square)

Therefore $\triangle OPA \cong \triangle OQC$ (S.A.S postulate)

(ii) $OP = OQ$ (radius)

$OC = OA$ (side of a square)

Therefore $OC - OP = OA - OQ$

$CP = QA$

Consider $\triangle BPC \ and \ \triangle BQA$

$BC = BA$ (side of a square)

$\angle PCB = \angle QAB = 90^o$

$CP = QA$

Therefore $\triangle BPC \cong \triangle BQA$

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Question 17: The length of common chord of two intersecting circles is $30 \ cm$. If the diameter of the circles be $50 \ cm$ and $34 \ cm$, calculate the distance between their center.

$OA = 17 \ cm$ and $O'A = 25 \ cm$ (given)

In $\triangle OAD \ \ OA^2=AD^2+OD^2$

$\Rightarrow OD^2 = OA^2-AD^2$

$OD^2 = 17^2 -15^2 = 289-225 = 64$

$\Rightarrow OD = 8 \ cm$

Similarly in $\triangle O'DA \ \ O'D^2 = O'A^2 - AD^2 = 25^2-15^2 = 400$

$\Rightarrow O'D = 20 \ cm$

Therefore $OO' = 8 + 20 = 28 \ cm$

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Question 18: The line joining the mid points of two chords if a circle passes through the center. Prove that the chords are parallel.

$AL = LB$ (given)

$OL$ bisects $AB$ (given)

Therefore $OL \perp AB$ (if line drawn from the center bisects the chord, then is is perpendicular to the chord)

Similarly $OM \perp CD$

Therefore $\angle DML = \angle BLP = 90^o$ (alternate angles)

Therefore $AB \parallel CD$

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Question 19: In the following figure, the line $ABCD \perp PQ$; where $P$ and $Q$ are the centers of the circles. Show that:

(i) $AB = CD$

(ii) $AC = BD$

$QO \perp AD \therefore AO = OD$ … … … … (i)

$PO \perp BC \therefore BO = CO$ … … … … (ii)

(i) – (ii) we get

$AO - BO = AD - CO \Rightarrow AB = CD$. Hence proved.

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Question 20: $AB$ and $CD$ are two equal chords of a circle with center $O$ which intersect each other at right angle at point $P$. If $OM \perp AB$ and $ON \perp CD$, show that $OMPN$ is a square.

$AB = CD$ and $AB \perp CD$ (given)

$OM \perp AB \Rightarrow AM = MB$

$ON \perp CD \Rightarrow CN = ND$

$\therefore all \angle \ in \ OMPN \ are 90^o$

$BM = \frac{1}{2} AB = \frac{1}{2} CD = CN$

Since $AB = CD$ and $OM = ON$ and because all angles are $90^o$ we can say

$OM = NP, ON = MP \Rightarrow NP = MP$

$MB = NC$

$MP + PB= NP+PC \Rightarrow PB = PC$

$\therefore CN = BM$

$CP+PN=PB+PM \Rightarrow PN = PM$

Therefore $OMPN$ is  square.

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Question 21: In the given figure, $O$ is the center of the circle. $AB$ and $CD$ are two chords of circle. $OM \perp AB$ and $ON \perp CD$. $AB = 24 \ cm, OM = 5 \ cm$, $ON = 12 \ cm$. Find the

(ii) length of chord $CD$ [2014]

$AB = 24 \ cm$, $OM = 5 \ cm$, $ON =12 \ cm$

$OM \perp AB$

$\Rightarrow AM = BM$

Therefore $AM = MB = 12 \ cm$

(i) Consider $\triangle AOM$

$AO^2 = AM^2 +OM^2 = 12^2+5^2 = 169$

$\therefore AO = 13 \ cm =$ Radius of the circle.

(ii) Now consider $\triangle CNO$

$CO^2 = CN^2+ON^2$

$\Rightarrow CN^2 = CO^2-ON^2 = 13^2 - 12^2 = 169 - 144 = 25$

$\therefore CN= 5 \ cm$