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# Circles : Exercise 10.2 (Mathematics NCERT Class 10th)

Q.1      From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(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, $OT \bot QT$
It is given that OQ = 25 cm and QT = 24 cm,
By Pythagoras theorem, we have $O{Q^2} = Q{T^2} + O{T^2}$
$\Rightarrow O{T^2} = O{Q^2} - Q{T^2}$
$\Rightarrow O{T^2} = {25^2} - {24^2}$
=(25+24)(25-24)
$= 49 \times 1 = 49$
$\Rightarrow OT = \sqrt {49} = 7$
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 $\angle POQ = {110^0}$, then $\angle PTQ$ is equal to
(A) ${60^0}$                 (B) ${70^0}$                  (C) ${80^0}$                      (D) ${90^0}$

Sol.       Since TP and TQ are tangents to a circle with centre O so that $\angle POQ = {110^0}$,
Therefore, $OP \bot PT$ and $OQ \bot QT$
$\Rightarrow \angle OPT = {90^0}$ and $\angle OQT = {90^0}$
In the quadrilateral TPOQ, we have
$\angle PTQ + \angle TPO + \angle POQ + \angle OQT = {360^0}$
$\Rightarrow \angle PTQ + {90^0} + {110^0} + {90^0} = {360^0}$
$\Rightarrow \angle PTQ + {290^0} = {360^0}$
$\Rightarrow \angle PTQ = {360^0} - {290^0}$
= ${70^0}$, 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 ${80^0}$, then $\angle POA$ is equal to
(A) ${50^0}$               (B) ${60^0}$                 (C) ${70^0}$                    (D) ${80^0}$
Sol.         Since PA and PB are tangents to a circle with centre O,

Therefore, $OA \bot AP,$ and $OB \bot BP$
$\Rightarrow \angle OAP = {90^0}\,\,and\,\,\angle OBP = {90^0}\,$
In the quadrilateral PAOB, we have
$\angle ABP + \angle PAO + \,\angle AOB + \angle OBP = {360^0}\,$
$\Rightarrow {80^0} + {90^0} + \angle AOB + {90^0} = {360^0}$
$\Rightarrow {260^0} + \angle AOB = {360^0}$
$\Rightarrow AOB = {360^0} - {260^0}$
=${100^0}$
In the right $\Delta s$ OAP and OBP, we have
OP = OP                           [Common]
OA = OB                          [Radii]
$\angle OAP = \angle OBP$   [Each = 90°]
Therefore, $\Delta OAP \cong \Delta OBP$   ( By SAS Criterion)
$\Rightarrow \angle POA = \angle POB$      [C.P.C.T.]
Therefore, $\angle POA = {1 \over 2}\angle AOB = {1 \over 2} \times {100^0}$
$= {50^0},i.e.,(A).$

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, $PQ \bot AB\,\,and\,\,PQ \bot CD$
$\Rightarrow \angle APQ = \angle PQD$
$\Rightarrow AB||CD$                     [Because $\angle APQ\,\,and\,\,\angle PQD$ 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, $AB \bot OP$ i.e., $\angle OPB = 90^\circ$
Also, $\angle QPB = 90^\circ$    (Construction)
Therefore, $\angle QPB = \angle OPB$, 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.

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, $\angle OTA = 90^\circ$
In right $\Delta OTA$, we have
$O{A^2} = O{T^2} + A{T^2}$
$\Rightarrow {5^2} = O{T^2} + {4^2}$
$\Rightarrow O{T^2} = {5^2} - {4^2} = 25 - 16 = 9$
$\Rightarrow OT = \sqrt 9 = 3$
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, $OP \bot AB$.
We know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.
So, $OP \bot AB$ and AP = BP.
In right $\Delta APO$, we have
$O{A^2} = A{P^2} + O{P^2}$
$\Rightarrow {5^2} = A{P^2} + {3^2}$
$\Rightarrow A{P^2} = {5^2} - {3^2} = 25 - 9 = 16$
$\Rightarrow AP = \sqrt {16} = 4$
Now, AB = 2AP                           [Because AP = BP]
$= 2 \times 4 = 8$
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)
$\Rightarrow AB + CD = BC + DA$
which proves the result.

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 $\angle AOB = 90^\circ$.
Sol.      Since tangents drawn from an external point to a circle are equal,
Therefore, AP = AC. Thus in $\Delta s$ PAO and AOC, we have
AP = AC
AO = AO [Common]
and, PO = OC [Radii of the same circle]
By SSS-criterion of congruence, we have
$\Delta PAO \cong \Delta AOC$
$\Rightarrow \angle PAO = \angle CAO$
$\Rightarrow \angle PAC = 2\angle CAO$
Similarly, we can prove that
$\angle QBO = \angle CBO$
$\Rightarrow \angle CBQ = 2\angle CBO$
Now, $\angle PAC + \angle CBQ = 180^\circ$ [Sum of the interior angles on the same side of transversal is 180°]
$\Rightarrow 2\angle CAO + 2\angle CBO = 180^\circ$
$\Rightarrow \angle CAO + \angle CBO = 90^\circ$
$\Rightarrow$   ${180^o} - \angle AOB = {90^o}$
[Since $\angle CAO,\,\angle CBO$  and $\angle AOB\,are\angle s$  of a triangle  Therefore $\angle CAO + \angle CBO + \angle AOB = {180^o}$]
$\Rightarrow \angle AOB = 90^\circ$

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-

$\angle AOB + \angle APB = 180^\circ$
In right $\Delta s$ 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
$\Delta AOP \cong \Delta OBP$
$\Rightarrow \angle OPA = \angle OPB$
and, $\angle AOP = \angle BOP \Rightarrow \angle APB = 2\angle OPA$
and, $\angle AOB = 2\angle AOP$
But $\angle AOP = 90^\circ - \angle OPA$              [Because $\Delta OAP$ is right triangle]
Therefore, $2\angle AOP = 180^\circ - 2\angle OPA$
$\Rightarrow \angle AOB = 180^\circ - \angle APB$
$\Rightarrow \angle AOB + \angle APB = 180^\circ$

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)
$\Rightarrow$ AB + CD = AD + BC
$\Rightarrow$ 2AB = 2BC [Because ABCD is a ||gm Therefore AB = CD and BC = AD]
$\Rightarrow$ 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 $\Delta ABC$ 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
$\Rightarrow$ s = 14 + x
${\rm{s - a = 14 + x - 14 = x,}}$
${\rm{s - b = 14 + x - x - 6 = 8}}$
${\rm{and, s - c = 14 + x - x - 8 = 6}}$
Therefore $Area\,\,(\Delta {\rm{ABC) = }}\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
${\rm{ = }}\sqrt {\left( {14 + x} \right)\left( x \right)\left( 8 \right)\left( 6 \right)}$
${\rm{ = }}\sqrt {48\left( {{x^2} + 14x} \right)}$
Also, Area $\left( {\Delta ABC} \right)$ = Area $\left( {\Delta OBC} \right)$ + Area $\left({\Delta OCA}\right)$ + Area  $\left( {\Delta OAB} \right)$
$= {1 \over 2} \times BC \times OD + {1 \over 2} \times CA \times OE + {1 \over 2} \times AB \times OF$
$= {1 \over 2} \times 14 \times 4 + {1 \over 2} \times (x + 6) \times 4 + {1 \over 2} \times (x + 8) \times 4$
$= 2\left( {14 + x + 6 + x + 8} \right)$
$= 2\left( {28 + 2x} \right)$
Therefore, $\sqrt {48\left( {{x^2} + 14x} \right)} = 2\left( {28 + 2x} \right),\,i.e.,\,4\left( {14 + x} \right)$
Squaring, we get
$48\left( {{x^2} + 14x} \right) = 16{\left( {14 + x} \right)^2}$
$\Rightarrow 3\left( {{x^2} + 14x} \right) = 196 + 28x + {x^2}$
$\Rightarrow 2{x^2} + 14x - 196 = 0$
$\Rightarrow {x^2} + 7x - 98 = 0$
$\Rightarrow \left( {x - 7} \right)\left( {x + 14} \right) = 0$
$\Rightarrow x = 7\,\,\,\,or\,\,\,\,\,x = - 14$
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-
$\angle AOB + \angle COD = 180^\circ$
and, $\angle AOD + \angle BOC = 180^\circ$
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, $\angle 1 = \angle 2,\angle 3 = \angle 4,\angle 5 = \angle 6\,\,\,and\,\,\angle 7 = \angle 8\,$
Now, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \,\angle 7 + \angle 8 = 360^\circ \,$
[Because Sum of all the $\angle s$ around a point is 360°]
$\Rightarrow 2\left( {\angle 2 + \angle 3 + \angle 6 + \angle 7} \right) = 360^\circ$
and, $2\left( {\angle 1 + \angle 8 + \angle 4 + \angle 5} \right) = 360^\circ$
$\Rightarrow \left( {\angle 2 + \angle 3 + \angle 6 + \angle 7} \right) = 180^\circ$
and $\left( {\angle 1 + \angle 8} \right) + \left( {\angle 4 + \angle 5} \right) = 180^\circ$
$\Rightarrow \angle AOB + \angle COD = 180^\circ$
$\Rightarrow [Since,\,\angle 2 + \angle 3 = \angle AOB,\angle 6 + \angle 7 = \angle COD\,\angle 1 + \angle 8 = \angle AOD\,\,and\,\,\angle 4 + \angle 5 = \angle BOC]$
and, $\angle AOD + \,\angle BOC = 180^\circ$

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