KN = KM + MN
But KM = KA and MN = BN, (\(\because\)  tangents drawn from an external point to a circle are equal)

KN = KA + BN
\(\therefore\)  Perimeter of \(\triangle PNK\) = PN + KN + PK
= PN + BN + KA + PK
= PB + PA = 2PB,
(\(\because\)  PA = PB)

Angle between the radii = 180°- 30° = 150°
(Since the sum of opposite angles = 180°).

5 cm

A line which intersects a circle in two distinct point is called a secant line.

10 cm

37

Given: A circle with the centre O. PA and PB are tangents to the circle form a point P.
To prove: (i) PAOB is a cyclic quadrilateral 
(ii) Po is the bisector of \(\angle \)APB
(iii) \(\angle \)OAB = \(\angle \)OPA
Proof: PA and PB are tangent to the circle at the point A and B respectively.
⇒ \(\angle \)OAP = \(\angle \)OBP = 90^o      .....(i)
[Tangent makes 90o angle with the radius at the point of contact]
⇒ \(\angle \)OAP + \(\angle \)OBP = 90^o + 90^o = 180^o       ......(ii)
In quadrilateral OAPB,
\(\angle \)AOB + \(\angle \)OAP + \(\angle \)OBP + \(\angle \)APB = 360^o
[Angle sum property of a quadrilateral]
⇒ \(\angle \)AOB + \(\angle \)APB + 180^o = 360^o  [From (ii)]
⇒ \(\angle \)APB + \(\angle \)AOB
= 360^o + 180^o = 180^o      ....(iii)
From (ii) and (iii), we have
\(\angle \)OAP + \(\angle \)OBP = 180^o
and \(\angle \)APB + \(\angle \)AOB = 180^o
⇒ Opposite angles of the quadrilateral are supplementary.
⇒ Quadrilateral OAPB is a cyclic quadrilateral.
(ii) In \(\triangle\)OAP and \(\triangle\)OBP
AP = BP [Tangent from the same external point are equal]
OP = OP    [common]
OA = OB   [Radii of the sama ecircle]
⇒ \(\triangle\)OAP ≅ \(\triangle\)OBP [Using SSS congruency]
⇒ \(\angle \)AOP = \(\angle \)BPO  [CPCT]    
⇒ PO bisects APB
(iii) \(\angle \)3 + \(\angle \)4 = 180^o     From (iii)
\(\angle \)1 + \(\angle \)2 + \(\angle \)3 = 180^o  [Angle sum property] ....(iv)
In \(\triangle\)OAB  OA = OB   [Radii]
⇒   1 = 2 [Angles opposite to equal sides of a \(\triangle\) are equal]    ......(v)
From (iii) and (iv)
\(\angle \)3 + \(\angle \)4 = \(\angle \)1 + \(\angle \)2 + \(\angle \)3
⇒ \(\angle \)4 = \(\angle \)1 + \(\angle \)2 = \(\angle \)1 + \(\angle \)1 = \(\angle \)2 + \(\angle \)2
\(\angle \)4 = 2\(\angle \)1 = 2\(\angle \)2 [From (v)]
⇒ \(1\over2\)\(\angle \)4 = \(\angle \)1  ⇒ \(\angle \)OPA = \(\angle \)OAB  Hence proved

Let AF = x cm
∵ AF = AE = x  [tangent from A]
Also BD = BF = 6 cm
and CD = CE = 9 cm
ஃ AB = (6 + x) cm and AC = (9+x)cm
Area \(\triangle\)ABC = Area \(\triangle\)BOC + Area \(\triangle\)COA + Area \(\triangle\)AOB
⇒ 54 = \(1\over2\) BC x OD + \(1\over2\)AC x OE + \(1\over2\)AB x OF
⇒ 54 x 2 = 15 x 3 + (6 + x) x 3 + (9 + x) x 3
108 = 45 + 18 + 3x + 27 + 3x
6x = 18 ⇒ x = 3
⇒ AB = 6 + x = 6 + 3 = 9 cm and 
AC = 9 + x = 9 + 3 = 12 cm

PA and PB are tangent to the circle
\(\angle \)A = \(\angle \)B = 90^o
[Tangent makes 90o angle with the radius at the point of contact]
In quadrilateral OAPB
\(\angle \)AOB + \(\angle \)A + \(\angle \)P + \(\angle \)B = 360^o
[Angle sum property of a quadrilateral].
⇒ \(\angle \)AOB + 90^o + 70^o + 90^o = 360^o
⇒ \(\angle \)AOB + 250^o = 360^o
⇒ \(\angle \)AOB = 360^o - 250^o = 110^o

In OAT \(\angle \)ATO = 40^o, \(\angle \)OAT = 90^o
∴ \(\angle \)AOT = 50^o [Angle sum property]
Now  \(\angle \)BTO = 40^o as OT bisects Similarly, \(\angle \)BOT = 50^o
ஃ \(\angle \)AOB = \(\angle \)AOT + \(\angle \)BOT = 50^o + 50^o = 100^o

In the figure given below, find \(\angle \)QSR. 
Given: PQ and PR are tangents to a circle with centre O and 
\(\angle \)QPR = 50
To find: QSR 
sol. \(\angle \)QOR + \(\angle \)QPR = 180°
⇒ \(\angle \)QOR + 50° = 180°
⇒ \(\angle \)QOR = 130QOR
⇒ \(\angle \)QOR = \(1\over2\) \(\angle \)QSR    [Degree measure theorem]
⇒  \(\angle \)QSR =  \(1\over2\) x 130° = 65°

Here OA = 6.5 cm and OC = 2.5 cm
\(\therefore \quad AC=\sqrt { { OA }^{ 2 }-{ OC }^{ 2 } } \)
\(=\sqrt { { \left( 6.5 \right) }^{ 2 }-{ \left( 2.5 \right) }^{ 2 } } \)
\(=\sqrt { 42.25-6.25}\) = 6 cm
\(\therefore\)  AB = 12 cm

Given: A circle C (O, r) with centre O. AB is a tangent to the circle at the point p. 
To Prove: OP ⊥ AB 
Const: Take any point Q on AB other than P and join OQ. 

OR > OP    .........(ii)
Thus, we find that among all such line segments, the line segment OP is the shortest, which is possible only when OP is perpendicular to AB. 
Hence, OP ⊥ AB. 

Construction: Join AD and BC
Proof: The tangent drawn from an internal point to a circle are equal in length.
If A is external point for circle hving centre O.
AB = AD    .....(i)
If C is external point then
BC = CD  .....(ii)
Now, B is external point for circle having centre O
AB = BC  .....(iii)
So, from (i), (ii) and (iii), we get
AB = BC = CD
So, AB = CD Hence proved.

Let AE = AF = x
Length of tangents from an external point are equal
ar \(\triangle\)BOC = \(1\over2\) x 7 x 2 = 7 cm2
ar \(\triangle\)BOC = \(1\over2\)x (4 + x) x2=(4 + x) cm2
ar  \(\triangle\)AOC = \(1\over2\) x (3 + x) x 2=(3  + x) cm2
ar \(\triangle\)ABC = \(1\over2\) AOB +ar \(\triangle\)BOC + a r\(\triangle\)AOC
S=\(a+b+c \over2\)
\(=\frac { 4+x+7+3+x }{ 2 } =\frac { 14+2x }{ 2 } =7+x\)
\(ar\Delta ABC=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { (7+x)(7+x-4-x)[7+x-7][7+x-3-x] } \)
\(=\sqrt { (7+x)\times 3\times x\times 4 } =2\sqrt { 3x(7+x) } \)
\(2\sqrt { 3x(7+x) } =4+x+7+3+x\Rightarrow 2\sqrt { 3x(7+x) } \) = 14 + 2x
\(\sqrt { 3x(7+x) } =7+x\Rightarrow 3x(7+x)={ (7+x) }^{ 2 }\)
⇒ 21x+3x² = 49 +x² + 14x ⇒ 2x² + 7x - 49 = 0
⇒ 2x² + 14x - 7x - 49 = 0 ⇒ 2x(x+7)-7(x+7) = 0
⇒ (2x - 7)(x + 7) = 0 ⇒ x = \(7\over2\), x = -7 [Rejected]
The length of side AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

Let P be any external point from which two tangents PA and PB are drawn to a circle with centre O.
To Prove: PA = PB
Proof : \(\angle OAP=\angle OBP=90\)
(radius is \(\bot\) to tangent)
In \(\triangle APO\) and \(\triangle BPO\),
\(\angle OAP=\angle OBP\)   (each 90°)
OA = OB  (radius)
OP = OP  common
\(\therefore \triangle APO\cong \triangle BPO\)  (by RHS)
\(\Rightarrow\)  PA = PB  (CPCT)

The distance between two parallel tangents is equal to diameter of the circle.
= 10 cm

Since, AD is a diameter of a circle, so AD is perpendicular to the tangent AB.
ㄥDAB =90°
In ΔABD,
ㄥDAB + ㄥABD +ㄥADB = 180°
90° + 50° + ㄥADB = 180°
ㄥADB = 180° -140° = 40°
In ΔODC,
OD = OC [same radii of circle]
ㄥOCD = ㄥCDO = 40°
ㄥDOC + ㄥOCD + ㄥCDO = 180°
ㄥDOC + 40° + 40° = 180°
ㄥDOC = 1,00°
Since, AD is a straight line.
ㄥDOC +ㄥCOA = 180°
100° +ㄥCOA = 180°
ㄥCOA = 80°

3 cm

(i)  AP = AR [Tangents from A] ...(i)
Similarly,  BP = BQ    ...(ii) 
CR = CQ   ...(iii) 
Now, ∵ AP = AR 
⇒ (AB - BP) = (AC-CR) 
⇒ AB + CR = AC+ BP 
⇒ AB + CQ = AC + BQ
(ii) Let AB = x, BC =y, AC = z 
ஃ Perimeter of \(\triangle\)ABC = x + y + z 
Area of \(\triangle\)ABC = [area of AOB + area of BOC + area AOC]
⇒ Area of ABC = AB x OP + x BC x OQ + x AC x OR
Area of ABC = \(1\over2\)X x r + \(1\over2\)y x r + \(1\over2\)z x 
\(\Rightarrow\) Area of \(\Delta\)ABC=\(1\over2\)(x+y+z) x r
\(\Rightarrow\) Area of \(\Delta\)ABC=\(1\over2\)(Perimeter of \(\Delta\)ABC) x r