
01. Real Numbers
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02. Polynomials
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03. Linear Equation
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04. Quadratic Equation
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05. Arithmetic Progressions
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06. Some Applications of Trigonometry
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07. Coordinate Geometry
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08. Triangles
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09. Circles
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10. Areas Related to Circles
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11. Introduction to Trigonometry
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12. Surface Areas and Volumes
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13. Statistics
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14. Probability
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15. Construction
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Lecture15.7

NCERT Solutions – Circles Exercise
Q.1 How many tangents can a circle have ?
Sol. A circle can have an infinite number of tangents.
Q.2 Fill in the blanks :
(i) A tangent to a circle intersects it in _________ point(s).
(ii) A line intersecting a circle in two points is called a _______.
(iii) A circle can have _______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ______.
Sol. (i) exactly one
(ii) secant
(iii) two
(iv) point of contact
Q.3 A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length of PQ is :
(A) 12 cm (B) 13 cm (C) 8.5 cm (D) 119−−−√cm
Sol. (D). Because,
PQ=OQ2−OP2−−−−−−−−−−√=122−52−−−−−−−√=144−25−−−−−−−√
=119−−−√cm
Q.4 Draw a circle and two lines parallel to a given line such that one is tangent and the other, a secant to the circle.
Sol. The required figure is as under :
Here ℓ is the given line and a circle with centre O is drawn.
Line PT is drawn  to line ℓ.
PT is the tangent to the circle.
AB is drawn  to line ℓ and is the secant.
(A) 7 cm (B)12 cm
(C) 15 cm (D) 24.5 cm
Sol. Since QT is a tangent to the circle at T and OT is radius,
Therefore, OT⊥QT
It is given that OQ = 25 cm and QT = 24 cm,
By Pythagoras theorem, we have
OQ2=QT2+OT2
⇒OT2=OQ2−QT2
⇒OT2=252−242
=(25+24)(2524)
=49×1=49
⇒OT=49−−√=7
Hence, radius of the circle is 7 cm , i.e., (A).
Q.2 In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ=1100, then ∠PTQ is equal to
(A) 600 (B) 700 (C) 800 (D) 900
Sol. Since TP and TQ are tangents to a circle with centre O so that ∠POQ=1100,
Therefore, OP⊥PT and OQ⊥QT
⇒∠OPT=900 and ∠OQT=900
In the quadrilateral TPOQ, we have
∠PTQ+∠TPO+∠POQ+∠OQT=3600
⇒∠PTQ+900+1100+900=3600
⇒∠PTQ+2900=3600
⇒∠PTQ=3600−2900
= 700, i.e. , (B).
Q.3 If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 800, then ∠POA is equal to
(A) 500 (B) 600 (C) 700 (D) 800
Sol. Since PA and PB are tangents to a circle with centre O,
Therefore, OA⊥AP, and OB⊥BP
⇒∠OAP=900and∠OBP=900
In the quadrilateral PAOB, we have
∠ABP+∠PAO+∠AOB+∠OBP=3600
⇒800+900+∠AOB+900=3600
⇒2600+∠AOB=3600
⇒AOB=3600−2600
=1000
In the right Δs OAP and OBP, we have
OP = OP [Common]
OA = OB [Radii]
∠OAP=∠OBP [Each = 90°]
Therefore, ΔOAP≅ΔOBP ( By SAS Criterion)
⇒∠POA=∠POB [C.P.C.T.]
Therefore, ∠POA=12∠AOB=12×1000
=500,i.e.,(A).
Q.4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Sol. Let PQ be a diameter of the given circle with centre O.
Let AB and CD be the tangents drawn to the circle at the end points of the diameter PQ respectively.
Since, tangent at a point to a circle is perpendicular to the radius through the point,
Therefore, PQ⊥ABandPQ⊥CD
⇒∠APQ=∠PQD
⇒ABCD [Because ∠APQand∠PQD are alternate angles]
Q.5 Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Sol. Let AB be the tangent drawn at the point P on the circle with O.
If possible, let PQ be perpendicular to AB, not passing through O.
Join OP.
Since, tangent at a point to a circle is perpendicular to the radius through the point,
Therefore, AB⊥OP i.e., ∠OPB=90∘
Also, ∠QPB=90∘ (Construction)
Therefore, ∠QPB=∠OPB, which is not possible as a part cannot be equal to whole.
Thus, it contradicts our supposition.
Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.
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Q.6 The Length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Sol. Since tangent to a circle is perpendicular to the radius through the point of contact,
Therefore, ∠OTA=90∘
In right ΔOTA, we have
OA2=OT2+AT2
⇒52=OT2+42
⇒OT2=52−42=25−16=9
⇒OT=9–√=3
Hence, radius of the circle is 3 cm.
Q.7 Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Sol. Let O be the common centre of two concentric circles, and let AB be a chord of the larger circle touching the smaller circle at P.
Join OP.
Since, OP is the radius of the smaller circle and AB is tangent to this circle at P,
Therefore, OP⊥AB.
We know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.
So, OP⊥AB and AP = BP.
In right ΔAPO, we have
OA2=AP2+OP2
⇒52=AP2+32
⇒AP2=52−32=25−9=16
⇒AP=16−−√=4
Now, AB = 2AP [Because AP = BP]
=2×4=8
Hence, the length of the chord of the larger circle which touches the smaller circle is 8 cm.
Q.8 A quadrilateral ABCD is drawn to circumscribe a circle (see figure).
Prove that AB + CD = AD + BC.
Sol. Let the quadrilateral ABCD be drawn to circumscribe a circle as shown in the figure.
i.e., the circle touches the sides AB, BC, CD and DA at P, Q, R and S respectively.
Since, lengths of two tangents drawn from an external point of circle are equal,
AP = AS
BP = BQ
DR = DS
CR = CQ
Adding these all, we get
(AP + BP) + (CR +RD) = (BQ+QC) + (DS+SA)
⇒AB+CD=BC+DA
which proves the result.
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Q.9 In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.
Prove that ∠AOB=90∘.
Sol. Since tangents drawn from an external point to a circle are equal,
Therefore, AP = AC. Thus in Δs PAO and AOC, we have
AP = AC
AO = AO [Common]
and, PO = OC [Radii of the same circle]
By SSScriterion of congruence, we have
ΔPAO≅ΔAOC
⇒∠PAO=∠CAO
⇒∠PAC=2∠CAO
Similarly, we can prove that
∠QBO=∠CBO
⇒∠CBQ=2∠CBO
Now, ∠PAC+∠CBQ=180∘ [Sum of the interior angles on the same side of transversal is 180°]
⇒2∠CAO+2∠CBO=180∘
⇒∠CAO+∠CBO=90∘
⇒ 180o−∠AOB=90o
[Since ∠CAO,∠CBO and ∠AOBare∠s of a triangle Therefore ∠CAO+∠CBO+∠AOB=180o]
⇒∠AOB=90∘
Q.10 Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Sol. Let PA and PB be two tangents drawn from an external point P to a circle with centre O. We have to prove that
∠AOB+∠APB=180∘
In right Δs OAP and OBP, we have
PA = PB [Tangents drawn from an external point are equal]
OA = OB [Radii of the same circle]
and, OP = OP [Common]
Therefore, by SSS – criterion of congruence
ΔAOP≅ΔOBP
⇒∠OPA=∠OPB
and, ∠AOP=∠BOP⇒∠APB=2∠OPA
and, ∠AOB=2∠AOP
But ∠AOP=90∘−∠OPA [Because ΔOAP is right triangle]
Therefore, 2∠AOP=180∘−2∠OPA
⇒∠AOB=180∘−∠APB
⇒∠AOB+∠APB=180∘
Q.11. Prove that the paralelogram circumscribing a circle is a rhombus.
Sol. Let ABCD be a parallelogram such that its sides touch a circle with centre O.
We know that the tangents to a circle from an exterior point are equal in length.
Therefore, AP = AS
BP = BQ
CR = CQ
and, DR = DS
Adding these, we get
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ 2AB = 2BC [Because ABCD is a gm Therefore AB = CD and BC = AD]
⇒ AB = BC
Thus, AB = BC = CD = AD
Hence, ABCD is a rhombus
Q.12 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.
Sol. Let the ΔABC be drawn to circumscribe a circle with centre O and radius 4 cm.
i.e., the circle touches the sides BC, CA and AB at D, E and F respectively.
It is given that BD = 8cm, CD = 6 cm.
Since, lengths of two tangents drawn from an external point of circle are equal.
Therefore, BF = BD = 8 cm,
CE = CD = 6 cm
and let AF = AE = x cm.
Then, the sides of the triangle are 14 cm, (x+6) cm and (x+8) cm.
Therefore, 2s = 14 + (x + 6) + (x+8) = 28 + 2x
⇒ s = 14 + x
s−a=14+x−14=x,
s−b=14+x−x−6=8
and,s−c=14+x−x−8=6
Therefore Area(ΔABC)=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
=(14+x)(x)(8)(6)−−−−−−−−−−−−−−−√
=48(x2+14x)−−−−−−−−−−−√
Also, Area (ΔABC) = Area (ΔOBC) + Area (ΔOCA) + Area (ΔOAB)
=12×BC×OD+12×CA×OE+12×AB×OF
=12×14×4+12×(x+6)×4+12×(x+8)×4
=2(14+x+6+x+8)
=2(28+2x)
Therefore, 48(x2+14x)−−−−−−−−−−−√=2(28+2x),i.e.,4(14+x)
Squaring, we get
48(x2+14x)=16(14+x)2
⇒3(x2+14x)=196+28x+x2
⇒2x2+14x−196=0
⇒x2+7x−98=0
⇒(x−7)(x+14)=0
⇒x=7orx=−14
But x cannot be negative,
Therefore, x = 7
Hence, the sides AB and AC are 15 cm and 13 cm respectively.
Q.13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Sol. Let a circle with centre O touch the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and s respectively.
We have to prove that
∠AOB+∠COD=180∘
and, ∠AOD+∠BOC=180∘
Join OP, OQ, OR and OS.
Since, the two tangents drawn from an external point to a circle subtend equal angles at the centre.
Therefore, ∠1=∠2,∠3=∠4,∠5=∠6and∠7=∠8
Now, ∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360∘
[Because Sum of all the ∠s around a point is 360°]
⇒2(∠2+∠3+∠6+∠7)=360∘
and, 2(∠1+∠8+∠4+∠5)=360∘
⇒(∠2+∠3+∠6+∠7)=180∘
and (∠1+∠8)+(∠4+∠5)=180∘
⇒∠AOB+∠COD=180∘
⇒[Since,∠2+∠3=∠AOB,∠6+∠7=∠COD∠1+∠8=∠AODand∠4+∠5=∠BOC]
and, ∠AOD+∠BOC=180∘
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