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Circles Ex- 10.2

 Hi Friends and Champs!
Before to proceed with the next exercise, we need to consider the following points.
* A point on the circle can have only and only one tangent.
* From a point outside the circle, we can only draw two tangents to the circle. 
* The length from a point to the point of contact is called length of the tangent.
* The length of two tangents drawn from outside to the circle is always equal. 
To prove the last point i.e. " the length of two tangents drawn from outside to the circle is always equal" , pls refer to the below figure:
Let a circle centre at "O" and "P" is a point outside the circle as shown in figure.
Construction: Draw PO and two tangent PQ and PR. Draw lines OQ and OR.

Since PQ and PR are tangents , therefore OQ and OR are perpendicular to the PQ and PR respectively. ( Theorem- 10.1 in previous chapter )
In triangle ΔPQO and ΔPRO
PO = PO is common 
OQ = OR ( radius of the circle )
<Q= <R = Right Angle 
=>ΔPQO ≅ ΔPRO   ( i.e. ΔPQO congruent to ΔPRO )
=>PQ = PR Hence proved.
Exercise 10.2
In Q.1 to 3, choose the correct option and give justification. 
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
Solution:
PQ = 24 cm
OQ = 25 cm
since OPQ is a right angle ( Theorem 10.1 ) , therefore
OQ2 = OP2 + PQ2
252 = OP2 + 242
625 - 576 = OP2
49 = OP2
OP = 7 cm Ans. option ( A) is correct
2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that < POQ = 110°, then <PTQ is equal to (A) 60° (B) 70° (C) 80° (C) 90°
Solution:

Construction: join T and O with line TO.
since the tangents form a point outside of the circle with remain equal , therefore
TP = TQ
TO = TO ( common )
PO = OQ ( radius)
therefore 
ΔTPO ≅  ΔTQP 
Hence <TOP = < TOQ = 110°/2 = 55
and <P = 70°
therefore <PTO = 180° - (90° +55°) 
<PTO = 180° - 145°
<PTO = 35° = <QTO    since ( ΔTPO ≅  ΔTQP )
Hence <T = 35° + 35° = 70° . Ans. option B is correct 
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°, then < POA is equal to 
 (A) 50° (B) 60° (C) 70° (D) 80
Solution:

Construction : Draw a line PO,
From theorem 10.1 , 
ΔPAO ≅ ΔPBO
=> <APO = 40°
since ΔPAO is a right angle triangle , right angle at P, therefore
<PAO = 180° - ( 90° + 40° )
<PAO = 50°Ans. option (A) is correct 
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Let PQ is the diameter and AB , CD are two tangents at point P and Q.
Since AB and CD are tangent therefore PO ⊥ AB and OQ ⊥ CD.
=> <APO = <DQO = 90°
and <CQO = <DQO = 90° 
since <APO and <DQO are alternate angles for the line AB and CD and both angles are equal to 90° , therefore AB and CD are parallel to each other. 
Ans. Hence proved
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:

Let PQ is the tangent on the circle with centre "O" and "Q" is the point of contact.
Since PQ is the tangent , therefore OQ ⊥ PQ, because tangent is always perpendicular to the radius.
Let R is a point within in the circle and QR ⊥ PQ .
but OQ ⊥ PQ , therefore
<PQO = <PQR = 90°
which is only possible when centre "O" and point "R" lying at the same line.
Hence it is proved that perpendicular at the point of contact is always passes thru centre of the circle.
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.
Solution:

OQ = 5 cm (given)
PQ = 4 cm (given)
OQ2 = PQ2 + PQ2
52 = 42 + PO2
PO2 = 25 - 16
PO2 = 9
PO = 3 Ans.
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.
Solution:
OP = 3 cm (given)  and OP ⊥ AB , since AB is the tangent and OP is the radius of small circle.
OB = 5 cm ( given )
since OPB is a right angle , therefore
OB2 = OP2 + PB2
52 = 32 + PB2
PB2 = 25 - 9
PB2 = 16
PB = 4
AP = PB ( since AB is the chord of  big circle and also AB is the tangent of small circle, therefore OP will bisect AB )
therefore, AB = AP + PB
AB = 4 + 4 = 8 cm Ans.
8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC
Solution:

Since PQRS are the points of contact , therefore,
AP =AS   ----- (1)     ( from Th 10.2)
BP = BQ  ----- (2)     ( from Th 10.2)
CR = CQ ----- (3)     ( from Th 10.2)
DR = DS ----- (4)     ( from Th 10.2)
From eq (1) + (4)
AP + DR = AS + DS
=> AP + DR = AD ----- (5)
From eq (2) + (3)
BP + CR = BQ + CQ
BP + CR = BC ----- (6)
From eq (5) + (6)
AP + DR + BP + CR = BC + AD
=> ( AP + BP ) + ( DR + CR ) = BC + AD
=> AB + CD = BC + AD  Ans. Hence proved
9. In Fig. 10.13, 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°.

Solution:
Construction: Join OC
Since XY , X'Y' and AB all are tangents, therefore OP ,OQ and OC all are perpendicular to the respective tangents.
Also OP =OQ=OC ( all are radius of the circle )
In ΔPOA and ΔCOA
AP = AC         ( both are tangents from a same point outside of the circle Th-10.2)
OP = OC         ( radius of same circle )
OA = OA         ( common side )
Hence from SSS congurency 
ΔPOA ≅ ΔCOA
therefore,
<POA = < COA  ----- (1)
similarly 
<QOB = < COB ----- (2)
from (1) + (2)
Since XY and X'Y' both are parallel to each other , therefore PQ will be diameter of the given circle,
hence 
<POA + <COA + <COB + <QOB = 180°   ( from eq (1) <POA = < COA and (2) <QOB = < COB )
=> <COA + <COA +<COB + <COB = 180°
=> 2 ( <COA + <COB ) = 180°
=> 2 <AOB = 180°
=> <AOB = 90° Ans. Hence proved
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.
Solution:
Let P is point outside of the circle and PQ , PR are the tangents touch at the point of contact Q and R.
In ΔPOQ and ΔPOR
PQ = PR ( Th 10.2)
PO = PO ( common side )
OQ = OR ( radius of the circle )
Hence by SSS congruency of triangle
ΔPOQ ≅ ΔPOR
Therefore , 
<QPO = <RPO           ( By CPCT  )
since ΔPOQ is a right angle, therefore 
<POQ = 90° - <QPO ------ (1)
similarly
<POR = 90° - RPO  -------(2)
from (1) + (2)
<POQ + <POR = 180° - (<QPO +<RPO )
<QOR = 180° - <QPR
=> <QOR + <QPR = 180°  Ans. Hence both angles are supplementary.
11. Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Solution: Let ABCD is a parallelogram with sides AB ∥ CD and BC ∥ DA. PQRS are the points of contacts.
AB = CD and BC = DA     ( since, ABCD is a parallelogram)
Since A,B,C,D are the points outside of the circle and AB,BC,CD,DA are the tangents , therefore from theorem 10.2
AP = AS  --------------(1)
BP = BQ --------------(2)
CR = CQ --------------(3)
DR = DS --------------(4)

since 
AB = AP + BP  and 
CD = CR + DR
then 
AB + CD = AP + BP + CR + DR ------- eq (5)
AB = CD,  since ABCD is a parallelogram. 
therefore, from eq (1),(2),(3) and (4), equation (5) could be written as
AB + AB = AS + BQ + CQ + DS            
2AB = ( AS +DS ) + ( BQ + CQ )
2AB = AD + BC
2AB = 2BC            ( AD = BC , since ABCD is a parallelogram )
AB = CD
Hence ABCD is a rhombus. Ans.
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 Fig. 10.14). Find the sides AB and AC.

Fig. 10.14
Solution:
Construction: Join OA,OB and OC

OD = 4 cm ( given)
CD = 6 cm ( given )
BD = 8 cm ( given )

Let x = AQ =  PQ            ( tangents are equal from a point outside of the circle ) 
CD = CP                          ( tangents are equal from a point outside of the circle ) 
BD = BQ                         ( tangents are equal from a point outside of the circle ) 

Total Area of ΔABC = area of ( ΔAPO + ΔAQO + ΔCOP + ΔCOD + ΔBOQ + ΔBOD ) --- (1)
                             
In ΔAPO and ΔAQO
AP = AQ ( tangents from a point to the circle )
OA = OA ( common )
OP = OQ ( radius)
therefore 
ΔAPO ≅  ΔAQO
similarly
ΔCOP ≅ ΔCOD and ΔBOQ ≅  ΔBOD
So from eq (1)  
Total area of ΔABC = 2 × area of ( ΔAPO + ΔCOP + ΔBOQ )
arΔABC = 2 × [ 2x + 12 + 16 ]           ( since area of triangle = 1/2 x Base x Height )
arΔABC = 4x + 56    -------------------------(2)

Now by heron's formula we already know 
arΔABC = √ [s(s - AB )( s -BC)(s-CA)]
where 
s = ( AB + BC + CA ) / 2
s = ( x + 8 + 14 + 6 + x) / 2
s = x + 14
hence 
arΔABC = √ [( x + 14 )(x+14 - x -8)(x+14 - 14)(x+14 -x-6)]
arΔABC = √[(x+14)(6)(x)(8)]
arΔABC= √48x(x+14)  ----------------------(3)
from eq (2) and (3)
4x + 56 = √48x(x+14)
square at both sides
16x² + 3136 + 448x = 48x² + 672x
32x² + 224x - 3136 = 0
x² + 7x - 98 = 0
x² + 14x - 7x - 98 =0
x(x + 14) -7( x + 14) =0
( x + 14 )(x-7) = 0
x = 7
so sides are 
AB = x+8 = 7+8 = 15 and
AC =x + 6 = 7 + 6 = 13
Ans.
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:

PQRS is the quadrilateral and A, B, C, D are the points of contacts of the circle inscribe in the quadrilateral.
then 
<POA = <POB  (since AP and AB are the tangents, therefore ΔPOA≅ ΔPOB ) 
similarly 
<SOC = <SOB
<COQ = <DOQ
<ROA = <ROD
But 
<POA +<POB +<SOC +<SOB + <COQ + <DOQ +<ROD +<ROA = 360 (complete angle of circle)
=>2 <POA + 2 <SOC + 2 <COQ + 2 <ROA = 360 
=> <POA + <SOC + <COQ + <ROA =180
=> ( <POA + <ROA ) + ( <SOC + <QOC ) =180
=> <POR + <SOQ = 180     ( since <POA + <ROA = <POR & <SOC + <QOC = <SOQ )
Therefore,  <POR and <SOQ angles subtended by opposite sides of the quadrilateral.
Hence proved.


I hope all the questions are properly readable and understandable, in case of some confusion, kindly let me know in comment section 🙏🙏🙏


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