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

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°

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.

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

We are given a circle with centre O and a tangent XY to the circle at a point P. We need to prove that OP is perpendicular to XY.

Take a point Q on XY other than P and join OQ see  Fig.The point Q must lie outside the circle. (Why? Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius OP of the circle. That is, OQ > OP.
Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY.

30\(°\)

\(In\ \Delta APB,\ \angle BAP=\angle ABP\)
(angle opp. to equal side)
\(\angle BAP={1\over2}(180-\angle APB)={1\over2}(130^0)=65^0\)|
\(\angle OAP =90^0\)
\(\angle OAP =90^0-\angle BAP=90^0-65^0=25^0\)

OAP = 90^o
⇒ 17² = r² + 15²    [By pythagoras theorem]
⇒ r² = 17² - 15² = (17 - 15)(17+15)
= 2 x  32
⇒ r² = 64 ⇒ r = \(\pm \) 8 cm
Ignoring negative value as length cannot be negative.
⇒ r = 8 cm

Given: TAS tangent to the circle with centre O at A.
\(\angle \)OBA = 32^o
To find: x
Sol. In \(\triangle\)OAB OA=OB      [RAdii of the same circle]
⇒   \(\angle \)1 = 32^o      [Angle opposite to equal sides of a triangle are equal]

In OAB
\(\angle \)1 + \(\angle \)2 + 32^o = 180^o     [Angle sum property of a triangle]
⇒ \(\angle \)2 + 32^o + 32^o = 180^o ⇒ <2 = 180^o - 64^o = 116^o
\(\angle \)3 = \(1\over2\)\(\angle \)2 = \(1\over2\) x 116^o = 58^o     
TAS is tangent to the circle (Given)
\(\angle \)x = \(\angle \)ABC = 58^o  [Angles in the alternate segments are equal]

PA = 12 cm, OA = 5 cm, OB = 3 cm
OP² = OA² + AP² = OB² + BP²
⇒ 25+14 = 9+BP2
⇒ 169-9 = BP²
⇒ BP = \(\sqrt{160}\) cm = 12.65 cm.(Approx.)

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°

Let O is the centre of the circle and P is a point such that OP = 25 cm and PQ is the tangent to the circle. 
OQ = radius = 7 cm
In \(\triangle\)OQP, we have \(\angle \)Q = 90°
OP² = OQ² + PQ²
⇒ (25)² = 7² + PQ² ⇒ PQ² = 625 - 49 = 576
⇒ PQ = 24 cm

64 cm

40 units

In right CAB
BC² = AC² + AB² = 8² + 6²
BC?² = 100 ⇒ BC = 10 cm

Area of \(\triangle\)CAB = \(1\over2\) x AB x AC
= \(1\over2\) x 6 x 8 = 24 cm²
Area of \(\triangle\)AOB  = \(1\over2\) x AB x OP
= \(1\over2\)x 6 x r = 3r cm²
Area of \(\triangle\)AOC = \(1\over2\) x AC x OT
= \(1\over2\) x 8 x r = 4r cm²
Area of \(\triangle\)BOC = \(1\over2\)x BC x OS
= \(1\over2\) x 10 x 10 x r = 5r cm²
Now, area \(\triangle\)AOB + area \(\triangle\)AOC + area \(\triangle\)BOC = Area \(\triangle\)ABC
⇒ 3r + 4r + 5r = 24 ⇒ 12r = 24
ஃ r = 2 cm

Given conditions are described in the diagram given below:

Now, as the perpendicular from centre to the chord of the same circle bisects that chord.
Hence, length of chord CD = 2EC = 2 x 4 = 8 cm

Let C₁ and C₂ be the two concentric circles, with centre at O and respective radii being r₁ = 4cm and r₂ = 5cm.
Draw a chord AC of circle C₂, to touch circle C₁ at B.
Join OB.

Here OB 丄 AC [As, tangent at any point of circle is perpendicular to radius through the point of contact]
Thus, in right angled OAB ,we have:
OA² = AB² + OB²     [By using phythagoras theorem]
⇒  52 =  AB² + 4²
⇒  AB² = 25 - 16  = 9
∴ Length of chord AC = 2 AB = 2 x 3 = 6 cm