We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P see fig. We are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then \(\angle\)OQP and \(\angle\)ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP,
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore,   \(\Delta\)OQP \(\cong\)\(\Delta\) ORP   (RHS)
This gives  PQ = PR    (CPCT)

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 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.

Join OB and OD, then OB = OD.

AB II PR 
\(\angle \)B = \(\angle \)Q [Alternate interior angles] 
⇒ \(\angle \)B = 70°
AQB is an Isosceles triangle, 
AB = AQ
So, \(\angle \)BQR = \(\angle \)BAQ
⇒ \(\angle \)A = 70
⇒ AQB
\(\angle \)A + \(\angle \)B + \(\angle \)Q = 180°
⇒ 70° + 70° + \(\angle \)Q = 180°
⇒ Q = 180° - \(\angle \)140° = 40°

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°

AB and CD are common tangents to the two given circles with centres O₁ and O₂. We know that lengths of the tangents drawn from a point outside the circle to the circle are equal in length.
\(\therefore\)  AE = EC and EB = ED
\(\Rightarrow\) AE + EB = CE + ED
\(\Rightarrow\) AB = CD.

Let AD = x cm; BD = AB - AD
= (12 - x)cm; AD = AF    [tangents from point A]
ஃ AF = x cm
Now, CF = AC - AF = (10-x) cm
and BD = BE ⇒ BE = (10 - x) c
Now, BC = CE + BE
⇒ 8 = (10 -x) + (12 - x)
⇒ 8 = 22 - 2x ⇒ 2x = 14
⇒ x = 7 cm
⇒ AD = 7 cm; BE = 12 - x
= 12 - 7 = 5 cm;
CF = 10 - x = 10 - 7 = 3 cm

Radius of bigger circle = \(1\over2\)d₂ and Radius of smaller circle = \(1\over2\)d₂
In right angled \(\triangle\)OAB
OB² = AB² + OA²
⇒ \({ \left( \frac { 1 }{ 2 } { d }_{ 2 } \right) }^{ 2 }={ \left( \frac { 1 }{ 2 } c \right) }^{ 2 }{ \left( \frac { 1 }{ 2 } { d }_{ 1 } \right) }^{ 2 }\)   [OA=\(1\over2\)d₁ and OB=\(1\over2\)d₂]
⇒ \(\frac { 1 }{ 4 } { d }_{ 2 }^{ 2 }=\frac { 1 }{ 4 } { c }^{ 2 }+\frac { 1 }{ 4 } { d }_{ 1 }^{ 2 }\)
⇒ \({ d }_{ 2 }^{ 2 }={ c }^{ 2 }+{ d }_{ 1 }^{ 2 }\)

Given. A circle C(O, r). PA and PB are tangent to the circle from point P, outside the circle such that \(\angle \)APB = 120^o. OP is joined.
To Prove. OP = 2AP.
Construction. Join OA and OB.
Proof. Consider s PAO and PBO
PA = PB    [Tangent to a circle, from a point outside it, are equal.]
OP = OP   [Common]
\(\angle \)OAP = \(\angle \)OBP = 90^o
ஃ  OAP OBP
ஃ  \(\angle \)OPA = \(\angle \)OPB = \(1\over2\) \(\angle \)APB = \(1\over2\) x 120^o = 60^o.
In right angled OAP, \(AP\over OP\) = cos60^o = \(1\over2\)⇒ OP = 2AP.

Given: AB, AC and AD are tangents. AB = 5 cm. 
TO find: AD 
Sol. AB and AC are tangents from the same point to the circle with centre O. 
⇒ AB = AC      .......(i)
(Length of the tangents from the same external point are equal). 
AC and AD are tangents from the same point to the circle with centre O. 
⇒ AC = AD    .....(ii)
(Length of the tangents from the same external point are equal) 
From (i) and (ii) 
∵ AB = AD = 5 cm 

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°

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 

Here AC is the diameter of the circle.
∴  ∠ABC = 90°     [ Angle in a semi-circle]
Now, in ΔACB, ∠A + ∠B + ∠C = 180°      [Sum of all interior angles of a triangle is 180°]
⟹     ∠A + 90° + 50° = 180°
⟹     ∠A + 140 = 180
⟹    ∠A = 180^o - 140° = 40°
Or     ∠OAB = 40°            …(i)
Here, OA ⏊ AT
⟹    ∠OAT = 90°          
⟹    ∠OAB + ∠BAT = 90°
⟹    ∠BAT = 90° - 40° = 50°    [Using (i)]
Hence, the value of ∠BAT is 50°.

(i) (b)
(ii) (c): In \(\Delta\)OAP and \(\Delta\)OBP,
\(\angle\)OAP = \(\angle\)OBP = 90°
[Since, radius at the point of contact is perpendicular to tangent]
OP = OP (Common)
OA = OB (Radii of circle)
So, \(\angle\)OAP == \(\angle\)OBP (By RHS congruency criterion)
(iii) (b): In quadrilateral OAPB, \(\angle\)AOB = 150°  [Given]
\(\angle\)OAP = \(\angle\)OBP = 90°
\(\therefore\) \(\angle\)APB = 360° - 90° - 90° - 150° = 30°
(iv) (d): We have, \(\angle\)APB = 40°

Now,PA =PB [Since, length of tangents drawn  from an external point are equal]
In \(\Delta\)PAB, \(\angle\)PAB = \(\angle\)PBA = 70° [Angles opposite to equal sides are equal]
Also, \(\angle\)PAB + \(\angle\)BAO = 90°
[Since, radius at the point of contact is perpendicular to tangent]
\(\Rightarrow\) \(\angle\)BAO = 90° - 70° = 20°
(v) (d): We have, \(\angle\)ABO = 45°

\(\because\) AO = OB (Radii of circle)
\(\therefore\) \(\angle\)BAO = \(\angle\)ABO = 45°  [Angles opposite to equal sides are equal]
Now, in \(\Delta\)OAB,
\(\angle\)AOB = 180° - 45° - 45° = 90°
Since, \(\angle\)APB = 360° - 90° - 90° - 90° = 90° i.e., AP\(\perp\) BP
So, OAPB is a square.

BQ = BP (length of tangents drawn from an external point to a circle are equal) ...(i)
CP = CR                                                                                                                ...(ii)
and AQ = AR
\(\therefore\) 2AQ = AQ + AR
= (AB + BQ) + (AC + CR)
= (AB + BP) + (AC + CP)
                  [using Eqs. (i) and (ii)]
\(\Rightarrow\)  2AQ = AB + AC + (BP + CP)
\(\Rightarrow\) 2AQ = AB + AC + BC
\(\begin{array}{ll} \Rightarrow & A Q=\frac{1}{2}(A B+B C+A C) \end{array}\)
\(\begin{array}{ll} \Rightarrow & A Q=\frac{1}{2} \text { Perimeter of } \triangle A B C \end{array}\)