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 $\triangle CNO$

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

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

$\therefore CN= 5 \ cm$

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