Let QS = x; SR = 12 - x
∴  PT = 5 - x; PM = PT
ஃ PM = 5 - x
Also   SR = MR ⇒ MR = 12 - x
Also  PQ² + QR² = PR²
⇒ PR = 13 ⇒ PM + MR = 13
⇒ 5 - x + 12 - x 13 ⇒ 2x  - 4 ⇒ x = 2
Also OSQT is a square
ஃ  OS = QS ⇒ OS = 2 cm
ஃ  Radius of incircle = 2 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

OP = Radius of the circle OA = 5 cm; AP = 4 cm
o [RAdius and tangent are ⊥ ar]
OA² = AP² + OP²   [By pythagoras theorem]
5² = 4² + OP²
⇒ 25 = 16 + OP²  ⇒ 25 - 16 = OP² ⇒ 9 = OP² ⇒ OP = \(\sqrt9\) = 3
Radius = 3 cm

Given: T BP and TCQ are tangents to the circle whose centre is O. 
Also,  \(\angle \)PBA = 60°
\(\angle \)ACQ = 70°

To determine: \(\angle \)BAC and \(\angle \)BTC
Sol. Join OB and OC
\(\angle \)OBP = 90°
[Tangent makes 90° angle with the radius at the point of contact]
⇒ \(\angle \)OBA + \(\angle \)ABP = 90°
⇒ \(\angle \)1 + 60° = 90° [Given ABP = 60°]
⇒ \(\angle \)1 = 30°       ......(i)
Also  \(\angle \)OCQ = 90°
⇒ OCA + 70° = 90°  ......(i)
⇒ \(\angle \)OCA = 20°   .....(ii)
In OBA ,   OB = OA = radii
⇒ \(\angle \)1 = \(\angle \)4 = \(\angle \)30°     .....(iii)
[ Angles opposite to equal sides of a triangle are equal]
Similarly, In \(\triangle\)OCA
OC = OA
\(\angle \) 5 = 20°​​​​​​​      .....(iv)
From (iii) and (iv)
\(\angle \)BAC = 20°​​​​​​​ + 30°​​​​​​​ = 50°​​​​​​​
⇒  \(\angle \)BOC = 2x50°​​​​​​​ = 100°​​​​​​​
\(\angle \)BOC + BTC = 180°​​​​​​​
100°​​​​​​​ + \(\angle \)BTC = 180°​​​​​​​
⇒ \(\angle \)BTC = 80°​​​​​​​

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. 

Let ABCD is a quadrilateral circumscribing a circle with centre O. Let circle touches the sides of a quadrilatcral at points E, F, G and H.

To prove \(\angle\)AOB + \(\angle\)COD = 180°
and \(\angle\) AOD + \(\angle\)BOC = 180°
Construction Join OE, OF, OG and OH.
Proof We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.
and       ....(i)
Also, we know that the sum of all angles subtended at a point is 360°.
\(\begin{array}{rlrl} \therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8 & =360^{\circ} \\ \end{array}\)    ...(ii)
\(\begin{array}{rlrl} \Rightarrow 2(\angle 2+\angle 3+\angle 6+\angle 7) & =360^{\circ} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & (\angle 2+\angle 3)+(\angle 6+\angle 7)=180^{\circ} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \angle A O B+\angle C O D=180^{\circ} \end{array}\)
Similarly, we have
\(\begin{aligned} 2(\angle 1+\angle 8+\angle 4+\angle 5) & =360^{\circ} \\ \end{aligned}\)  [from Eq. (i) and (ii)]
\(\begin{aligned} & \Rightarrow(\angle 1+\angle 8)+(\angle 4+\angle 5)=180^{\circ} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \angle A O D+\angle B O C=180^{\circ} \end{aligned}\) Hence proved.

\(\angle QPR={ 50 }^{ ° }\)          (Given)
\(\therefore \quad \angle QOR={ 180 }^{ ° }-{ 50 }^{ ° }={ 130 }^{ ° }\)
From \(\triangle OQR\)
\(\Rightarrow \quad \angle QOR=\angle ORQ=\frac { { 180 }^{ ° }-{ 130 }^{ ° } }{ 2 } \)
\(=\frac { { 50 }^{ ° } }{ 2 } ={ 25 }^{ ° }\)

\(\angle OAC={ 90 }^{ ° }\)  (as radius \(\bot\) r tangent)
\(\angle BOC=\angle OAC+\angle ACO\)
(Exterior angle property)
\(\Rightarrow \quad { 130 }^{ ° }={ 90 }^{ ° }+\angle ACO\)
\(\Rightarrow \quad \angle ACO={ 130 }^{ ° }+{ 90 }^{ ° }\)
\({ =40 }^{ ° }\)

Given \(\angle \)APB = 60°
∵  \(\angle \)APB + \(\angle \)PAB + \(\angle \)PBA = 180°   [∵ PA = PB ⇒ \(\angle \)PAB = \(\angle \)PBA]
⇒ \(\angle \)APB + x + x = 180°
⇒ APB + 2x = 180°
⇒ 2x = 180° - 60°
⇒ 2x = 120° ⇒ x = 120°/2 = 60°
Also \(\angle \)OAP = 90°
⇒ \(\angle \)OAB + \(\angle \)PAB = 90°
\(\angle \)OAB + 60° = 90° ⇒ \(\angle \)OAB = 30°

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°

I common tangent can be drawn as shown in the figure. 

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