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)

Given, DA and DC are the tangents of a circle and LADC =50°.
We know that, radius of a circle is perpendicular to the tangent at the circle.
OA丄AD and OC丄CD.

ㄥOAD=ㄥOCD=90^o
ㄥOAD+ㄥOCD = 90° + 90° = 180°
OADCis a cyclic quadrilateral.
ㄥADC+ㄥAOC = 180°
50° + ㄥAOC = 180°
ㄥAOC = 180°- 50° = 130°

19.5 cm and 22.5 cm

AD = CD
ஃ AD = 6 cm
Also DB = CD 
ஃ DB = 6 cm
⇒ AB = 6 + 6
=- 12 cm

If we draw tangent from an external point to a circle then a line joining the given point and centre of a circle always bisect the angle formed by these two tangents. 

i.e., \(\angle \)1 = \(\angle \)2
From above property
In given figure
\(\angle \)1 = \(\angle \)2
\(\angle \)3 = \(\angle \)4
\(\angle \)AED = \(\angle \)CEB (V.O.A)   .......(i)
So,  1 + 2 + CEB + 4 + 3 + AED = 360° [Angles on a point]
From (i),
\(\angle \)1 + \(\angle \)1 + \(\angle \)AED + \(\angle \)3 + \(\angle \)3 + \(\angle \)AED = 360°
⇒ 2\(\angle \)1 + 2\(\angle \)AED + 2\(\angle \)3 = 360°
⇒ \(\angle \)1 + \(\angle \)AED + \(\angle \)3 = 180°
The sum of angles on a straight line is 180°
It shows OEO' are collinear.

OP is tangent of the circle having centre O'
So \(\angle \)OPO' = 90°
[ ∵ Radius and tangent are to ⊥ each other at the point of contact]
In right-angled OPO'
OP = 4 cm
O'P = 3 cm  [Given]
In right-angled \(\triangle\)OPO'
OP = 4 cm
O'P = 3 cm   [Given]
OO'² = OP² + O'P²
= 4² + 3² = 16 + 9 = 25
OO' = 5 cm
If two circles intersect each other then line joining the two centre always ⊥ bisector of the common chord.
OO' ⊥ PQ and PT = TQ
Area of \(\triangle\) OO'P = \(1\over2\) x base x altitude
Here, base = 4 cm, attitude = 3 cm.
Area = \(1\over2\) x 4 x 3 = 6 cm²   ...(i)
But if base OO' = 5 cm altitude = PT
Area \(\triangle\)POO' = \(1\over2\) x 5 x altitude   .....(ii)
6 cm2 = \(1\over2\)x 5 x altitude  .....(ii)
Comparing (i) and (ii)
6 cm² = \(1\over2\) x 5 x altitude
⇒ \(\frac { 2\times 6 }{ 5 } \) = Altitude
⇒ \(\frac { 12 }{ 5 } \) = PT
⇒ PQ = 2PT = \(\frac { 2\times 12 }{ 5 } =\frac { 24 }{ 5 } \)cm
So, length of common chord = \(24\over5\) cm
= 4.8 cm

Given: Chord AB = 24 cm, radius 0B = OA = 13 cm. 
Construction: Draw OP ⊥ AB 
sol: In \(\triangle\)OPB 
OP ⊥ AB 
⇒ AP = PB [Perpendicular from centre on chord bisect the chord] 
= \(1\over2\)AB = 12 
OB² = OP² = PB²
⇒ (13)² = OP² + PB² ⇒ 169 = OP² + (12)²
OP2 = 169 - 144 = 25 ⇒ OP = 5 cm
In \(\triangle\)BPC   BC² = x² + k²
BC² = x² + 144         ..........(i)
In \(\triangle\)OBC  OC² = OB² + BC²  
(x + 5)² = (13)² + BC²
⇒ x² + 25 + 10x = 169 + BC²
⇒ x² + 25 + 10x = 169 + x² + 144
⇒ 25 + 10x = 169 + 144
⇒ 10x = 169+ + 144 - 25 = 288
⇒ x = \(288\over10\) = 28.8 cm
Put value of x in (i) BC2 = x2 + 144 = \(\frac { { (144) }^{ 2 } }{ 25 } \) + 144 = 144\(\left( \frac { 144 }{ 25 } +1 \right) \) = \(\frac { 144(169) }{ 25 } \)
BC = \(\frac { 12\times 13 }{ 5 } =\frac { 156 }{ 5 } =31.2\)
AC = BC = 31.2 cm

Join OA and OB. 
In \(\triangle\)PAO and \(\triangle\)PBO 
OA = OB    [Radii] 
OP = OP   [Common]
and AP = BP   [Tangent from P]
ஃ  \(\triangle\)PAO ≌ \(\triangle\)PBO   (SSS) 
⇒ ∠1 = ∠2
In \(\triangle\)ACP = \(\triangle\)BCP 
\(​​​​\angle \)1 = ∠2  [proved]
AP = BP and PC = PC
\(\triangle\)APC ≅ \(\triangle\)BPC    [SAS]
AC = BC  [CTCT]
and \(\angle \)ACP = \(\angle \)BCP
Also, \(\angle \)ACP + \(\angle \)BCP = 180° ⇒ ACP = 90°

In \(\triangle\)AOC, OA = OC [Radii of the same circle]
∴ \(\angle \)ACO = \(\angle \)CAO = 30° [Opp Also angles of equal sides are equal.]
Also \(\angle \)ACB = 90°  [Angle in semicircle]
∴ \(\angle \)OCB = 90° - 30° = 60°
In \(\triangle\)COB, OC = OB   [Radii of the semicircle]
∴ \(\angle \)COB = \(\angle \)OBC = 60°  [Opposite angles of equal sides]
Now OC ⊥ CQ
ஃ \(\angle \)OCQ = 90° ⇒  \(\angle \)BCQ = 90° - 60° = 30°
Also \(\angle \)OBC + \(\angle \)CBQ = 180°​
⇒  60°+\(\angle \)CBQ = 180° ⇒ CBQ = 120°
In \(\triangle\)CBQ
\(\angle \)BCQ + \(\angle \)CBQ + \(\angle \)CQB = 180°
⇒ 30° + 120° + CQB = 180° ⇒ CQB = 30°
\(\angle \)CQA = 30°
\(\angle \)CBA = 180° - \(\angle \)CBQ = 180° - 120° = 60°