Class 10: Circles – Exercise 18(a) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 $AB$ .

Answer:

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

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

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]

Answer:

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

Answer:

(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.

Answer:

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) $CB$

Answer:

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

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

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

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

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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.

Answer:

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 $AB = CD$ .

Answer:

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.

Answer:

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) $\angle AMN = \angle CNM$

c13 Answer:

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) $AB = CD$

c14 Answer:

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

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

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

Answer:

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) $\triangle BPC \cong \triangle BQA$

c16 Answer:

(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.

c17 Answer:

$OA = 17 \ cm$ 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.

Answer:

$AL = LB$ (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) $AC = BD$

Answer:

$QO \perp AD \therefore AO = OD$ … … … … (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.

c19 Answer:

$AB = CD$ 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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c11 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