Let C₁ and C₂ be rwo circles of radii, r₁=3 cm and r₂ = 5 cm and having common centre O.
Now, let AB be the chord of circle C₂ such that it touches the circle C₁ at point D. Clearly, AB be the tangent to the circle C₁, at point D.

\(\therefore\) OD \(\perp\)AB   [\(\because\) radius is perpendicular to the tangent at the point of contact]
\(\Rightarrow\) AD = BD [ \(\because\) perpendicular from centre to the chord bisect the chord]
Now, in right angled \(\Delta\)ODB,
OB² = OD² + DB²   [by Pythagoras theorem]
\(\Rightarrow\) 5² = 3² + DB² \(\Rightarrow\) DB² = 25 - 9 = 16
\(\Rightarrow\) DB = 4 cm [ taking positive square root]
\(\therefore\) Length of chord = AB = 2 AD = 2 \(\times\)4 = 8 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.

10 cm

From every point of a circle, we can draw a tengent. Therefore, infinite tangents can be drawn.

4 cm

5 cm

\(AB=2\sqrt{2r_1r_2-r^{2}_{2}}\)

Given: Two circles C₁ and C₂ touch each other at P where C₁ be inside of C₂. T be a point on the tangent at P, where TQ and TR are the tangents, segments to the circles. 
To prove: TQ = TR 
Proof: In circle C₁, TP and T Q are two tangents from an external point T. Then they are equal in length. 
I.e.,  TP = TQ ...(i)
Similarly, in circle C₂, TP = TR  .....(ii)
From (i) and (ii), we get 
TQ = TR Proved.

Given: AB is arc in circle C(O, r), P is mid-point of arc AB; XY is tangent to the circle at P.
To prove: AB | | XY. Join OA and OB.
Here  \(\angle \)AOP = \(\angle \)BOP (angle subtended by equal are)
OA = OB, OC = OC
∴  \(\triangle\)ACO ≅ \(\triangle\)BCO   (SAS)
⇒  AC = BC
⇒ OC ⊥ AB (line joining mid-point of chord with centre of circle is perpendicular to the chord)
Also  \(\angle \)OPY = 90°
⇒  \(\angle \)OCB = \(\angle \)OPY  (corresponding angles)
∴  AB | | XY

To prove: BQ = QC

AB is diameter
ஃ  \(\angle \)APB = 90^o  [Angle is semicircle]
⇒  BP ⊥ AC
⇒ \(\angle \)3 + \(\angle \)4 = 90^o   .......(i)
BQ = QP  [Tangents from Q]  .......(ii)
ஃ   \(\angle \)3 = \(\angle \)1

ஃ  PR and RC are tangent to circle with centre A.

PR=RC [Tangent from point R].....(i)
Similarly, RQ and RC are tangent to circle with centre B
∴ RQ = RC ....(ii)
From (i) and (ii), PR = RQ
ஃ R bisects PQ.

PA and PB are tangent to the circle
\(\angle \)A = \(\angle \)B = 90^o
[Tangent makes 90o angle with the radius at the point of contact]
In quadrilateral OAPB
\(\angle \)AOB + \(\angle \)A + \(\angle \)P + \(\angle \)B = 360^o
[Angle sum property of a quadrilateral].
⇒ \(\angle \)AOB + 90^o + 70^o + 90^o = 360^o
⇒ \(\angle \)AOB + 250^o = 360^o
⇒ \(\angle \)AOB = 360^o - 250^o = 110^o

Join OL
OL RS
Also    OL  = OM   [Radii of the same circle]
ஃ ⇒ OLM = 30^o
⇒ o - 30^o = 60^o

AR = 4cm
Also, AR = AQ ⇒ AQ = 4cm
Now, QC = AC - AQ
= 11 cm - 4 cm = 7 cm
Also, BP = BR
ஃ  BP = 3 cm and PC = QC
ஃ  PC = 7 cm  [From (i)]
BC = BP + PC = 3cm = 10 cm