Let circle touch CB at M and CA at N, AB at P
Now OM \(\bot\) CB and ON \(\bot\)AC (radius \(\bot\) tangent)
OMCN is square
Let OM = r = CM = CN
AN = AP, CN = CM, BM = BP
(tangant from external point)
AN = AP
or AC - CN = AB - BP
b - r = c - BM
b - r = c - (a - r)
b - r = c - a + r
2r = a + b - c
\(r=\frac { a+b-c }{ 2 } \)

Let PL = y, OP is \(\bot\) bisector of AB
\(\Rightarrow\)  AL = BL = 8 cm
OL² = OA² - AL² = 102 - 82 = 36
\(\Rightarrow\)  OL = 6cm
In \(\triangle OAP\),  AP²  = (y + 6)² - (10)² ....(i)
In \(\triangle ALP\),  AP² = y² + 64 ...(ii)
From (i) and (ii)  \(y=\frac{32}{3}\)
\(\therefore AP=\frac{40}{3}\) 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

Let PQ and PR be two tangents drawn from an external point P to a circle with centre O.

To prove \(\angle\)QOR = 180° – \(\angle\)QPR
or \(\angle\)QOR + \(\angle\)QPR = 180°
Proof In \(\Delta\)OQP and \(\Delta\)ORP,
PQ = PR   [\(\because\) tangents drawn from an external point are equal in length]
OQ = OR      [radii of circle]
OP  = OP   [common sides]
\(\therefore\)  \(\Delta\)OQP \(\cong\) \(\Delta\)ORP   [by SSS congruence rule]
Then, \(\angle\)QPO = \(\angle\)RPO    [by CPCT]
and \(\angle\)POQ = \(\angle\)POR    [by CPCT]
\(\left.\begin{array}{ll} \Rightarrow & \angle Q P R=2 \angle O P Q \\ \text { and } & \angle Q O R=2 \angle P O Q \end{array}\right\}\)    ...(i)
Now, in right angled \(\Delta\)OQP, \(\angle\)QPO + \(\angle\)QOP = 90°
\(\Rightarrow\)  \(\angle\)QOP = 90° - \(\angle\)QPO
\(\Rightarrow\)  2\(\angle\)QOP = 180°- 2 \(\angle\)QPO
[multiplying both sides by 2]
\(\Rightarrow\) \(\angle\)QOR = 180° - \(\angle\)QPR      [from Eq. (i)]
\(\Rightarrow\)   \(\angle\)QOR + \(\angle\)QPR = 180°      Hence proved.

QP = 3.8
QP = PT
(Length of tangents from external points are equal)
\(\Rightarrow\)  PT = 3.8 cm
PR = PT = 3.8 cm
\(\Rightarrow\)  QR = 7.6 cm

Let radius of circle = R
∴  \(\pi\)R2 = 25\(\pi\)
⇒ R = 5 cm
Distance between parallel tangents 
= 2 x 10 cm. 

\(\angle \)QPA = \(\angle \)RPA 

⇒ \(\angle \)RPA = 27°
\(\angle \)QPR = \(\angle \)QPA + RPA 
= 27° + 27° = 54°
Now,  \(\angle \)QAR + \(\angle \)QPR = 180°
⇒ \(\angle \)QAR = 180° - 54° = 126°

\(\angle \)AOB = 100°
\(\angle \)AOB + \(\angle \)OAB + \(\angle \)OBA = 180°
⇒ 100° + x + x = 180° [Let \(\angle \)OAB = x = \(\angle \)OBA]
angle opp. to equal side
⇒ 2x = 180° - 100°
⇒ 2x = 80° ⇒ x = 40°
\(\angle \)OAB + \(\angle \)BAT = 90°​​​​​​​
⇒ ​​​​​\({ 40 }^{ \underset { \_ }{ O } }\) + BAT = 90°
⇒ \(\angle \)BAT = 50 °​​​​

Given: PQ is a tangent to the circle with centre O from a point P. QCR is a diameter of the circle
and \(\angle \)POR = 110°. 
To find: \(\angle \)OPQ 

Sol. POR = 110°
QR is the diameter of the circle.
⇒ \(\angle \)1 + \(\angle \)2 = 180°  [Linear pair axiom]
⇒ \(\angle \)1 + 110° = 180° ⇒ \(\angle \)1 = 70°
\(\angle \)OQP = 90°
(Tangent makes 90° angle with the radius at the point of contact).
In \(\triangle\)OPQ
\(\angle \)1 + \(\angle \)OQP + \(\angle \)QPO = 180°      [Angle sum properety]
⇒ 70° + 90° + \(\angle \)QOP = 180°
⇒  \(\angle \)OPQ = 180° - 160°
⇒ \(\angle \)OPQ = 20°

From an external point P only two tangents can be drawn to a  circle.
Value of k = 2.

parallel tangents of a circle can be drawn only at the end points of the diameter 
⇒  I₁ || I₂
⇒  Distance between I₁ and I₂ = AB = Diameter of the circle 
= 2 x r = 2 x 7 cm = 14 cm 

Given: A circle with centre O. AB is the diameter of this circle. I is tangent to the circle. AH and BK are perpendicular to I from A and B at H  and K respectively. 
To prove: AH + BK = AB 
Proof: AH and HP are tangents from the external point H 
ஃ  AH = HP        .....(i)
and BK, KP are tangent from the external point K
ஃ   BK = KP      ......(ii)
Adding (i) and (ii) we get
AH + BK = HP + PK = HK ......(iii)
AB ⊥  AH
AB ⊥ BK
[Tangent makes 90° angle with radius at the point of contact]
⇒ \(\angle \)1 = \(\angle \)2 = 90°
Given that AH ⊥ i ⇒ \(\angle \)3 = 90°
and BK ⊥ i ⇒ \(\angle \)4 = 90°
∵ \(\angle \)1 = \(\angle \)2 = \(\angle \)3 = \(\angle \)4 = 90°​​​​​​​
⇒ AHKB is a rectangle
⇒  AB = HK        ......(iv)
[Opposite sides of a rectangle are equal]
From (iii) and (iv) ⇒ PH + PK = AB
AH + BK = AB from (i) and (ii) Hence proved.

OQ 1 AB l [Radius is perpendicular to the tangent) OP 1 BC 1 
ஃ   OPBQ is a square.
⇒ BQ = BP = OP = r.
Now  RD = DS ⇒ RD = 5 cm
ஃ   AR = AD - RD = 23 - 5 = 18 cm
Also,  AR = AQ ⇒ AQ = 18 cm
Now, AB = AQ + BQ ⇒ 29 = 18 + r ⇒ r = 11 cm.