Given, CD = 6 cm, BD = 8 cm and radius = 4 cm

Join OC, OA and OB.
Let the circle touches the other sides AB and AC at points E and F, respectively.
We know that tangents drawn from an external point to the circle are equal in length.
\(\therefore\) CD = CF = 6 cm            [\(\because\) C is an external point]
BD = BE = 8 cm                [\(\because\) B is an external point]
and AF = AE = x cm (say  [\(\because\) A is an external point]
Area of \(\Delta\)OCB, \(A_1=\frac{1}{2} \times \text { Base } \times \text { Height }\)
\(\begin{aligned} & =\frac{1}{2} \times C B \times O D \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 14 \times 4=28 \mathrm{~cm}^2 \\ \end{aligned}\)
\(\begin{aligned} & \quad[\because C B=C D+B D=6+8=14] \end{aligned}\)
Area of \(\Delta\)OCA,
\(\begin{aligned} A_2 & =\frac{1}{2} \times A C \times O F \end{aligned}\)
\(\begin{aligned} =\frac{1}{2}(6+x) \times 4=(12+2 x) \mathrm{cm}^2 \end{aligned}\)
and area of \(\Delta\)OBA,
\(A_3=\frac{1}{2} \times A B \times O E=\frac{1}{2}(8+x) \times 4=(16+2 x) \mathrm{cm}^2\)
Thus, area of  \(\Delta\)ABC
= A₁ + A₂ + A₃ = [28 + (12 + 2x) + (16+ 2x)]
= (56 + 4x) cm²     ...(i)
Now, semi-perimeter of \(\Delta\)ABC=\(\frac{1}{2}\)(AB + BC + CA)
\(\Rightarrow \quad s=\frac{1}{2}(x+8+14+6+x)\)
\(\Rightarrow\) s = (14 + x) cm
Using Heron's formula,
area of \(\Delta\)ABC = \(\begin{aligned} & =\sqrt{s(s-a)(s-b)(s-c)} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x)(14+x-14)(14+x-x-6)(14+x-x-8)} \end{aligned}\)
\(\begin{aligned} & =\sqrt{(14+x) \times x \times 8 \times 6} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x) 48 x} \end{aligned}\)     ...(ii)
From Eqs. (i) and (ii), we get
\(\sqrt{(14+x) 48 x}=56+4 x=4(14+x)\)
On squaring both sides, we get
(14 + x) 48 x = 4² (14 + x)²
\(\Rightarrow\) 3x = 14 + x
\(\Rightarrow\) 2x = 14
\(\Rightarrow\) x = 7
\(\therefore\)  Length of AC = 6 + x = 6 + 7 = 13 cm
and length of AB = 8 + x = 8 + 7 = 15 cm

Given XY and X'Y' are two parallel tangents. Another tangent AB touches the circle at C and intersect XY at A and X'Y'at B.
To prove \(\angle\)AOB = 90°
Proof We know that, tangents drawn from an external point to a circle are equal in length.
\(\therefore\)   AP = AC   [\(\because\) A is an external point] ...(i)
Thus, in \(\Delta\)APO and \(\Delta\)ACO, AP = AC   [from Eq. (i)]
AO = AO  [common sides]
OP = OC [radii of circle]
\(\Delta\)APO \(\cong\)\(\Delta\)ACO  [by SSS congruence rule]
Then, \(\angle\)OAP = \(\angle\)OAC    [by CPCT]        ...(ii)
\(\Rightarrow\) \(\angle\)PAC = 2 \(\angle\)CAO                             ....(iii)
Similarly, we can prove that \(\angle\)CBO = \(\angle\)OBQ
\(\Rightarrow\)   \(\angle\)CBQ = 2 \(\angle\)CBO                           ...(iv)
since, XY || X'Y'      [given]
\(\therefore\) \(\angle\)PAC +  \(\angle\)QBC = 180°
[\(\because\) sum of interior angles on the same side of transversal is 180°]
\(\Rightarrow\) 2 \(\angle\)CAO + 2 \(\angle\)CBO = 180°    [from Eqs. (iii) and (iv)]
\(\Rightarrow\)   \(\angle\)CAO + \(\angle\)CBO = 90°                    ...(v)
Now, in \(\Delta\)AOB, \(\angle\)CAO + \(\angle\) CBO + \(\angle\)AOB = 180°
[by angle sum property of triangle]
\(\Rightarrow\) \(\angle\)CAO + \(\angle\)CBO = 180° - \(\angle\)AOB    ...(vi)
From Eqs. (v) and (vi), we get
180° - \(\angle\)AOB = 90° \(\Rightarrow\) \(\angle\)AOB = 90° Hence proved.

Given A quadrilateral ABCD, circumscribing a circde.
To prove AB + CD = AD + BC
Proof We know that the lengths of tangents drawn from an external point to a circde are equal.
\(\therefore\)   AP = AS               ...(i)
[\(\because\) both are tangents to a circle from point A]
Similarly, BP = BQ,   ...(ii)
CR = CQ                  ....(iii)
and DR = DS           .....(iv)
On adding Eqs. (i), (ii), (ii) and (iv), we get
(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)
\(\Rightarrow\)   AB + CD = (AS + DS) + (BQ + CQ)
\(\Rightarrow\)  AB + CD = AD + BC
Hence Proved.

Given conditions are described in the diagram given below:

We are to find the area of the quadrilateral PQRL
Here OQ⊥QP [As, tangent at any point of circle is perpendicular to radius through the point of contact]
Similarly, OR⊥PR
Now in ΔOPQ and ΔOPR
OP = OP [Common Side]
ㄥQ =  ㄥR = 90^o
And, ΔOPQ  ≌ ΔOPR
Area of quadrilateral PQOR = 2 X area of ΔOQP
In right angled OQP,
OP² = OQ² + PQ²
⇒  132 = 5² + PQ²
⇒  PQ² = 169 - 25  =144
⇒  PQ  = 12 cm
Now, area of OQP = \(\frac { 1 }{ 2 } \times PQ\times OQ\)
=  \(\frac { 1 }{ 2 } \times 12\times 5\) = 30 cm²
∴ Area of quadrilateral QORP = 2 x area of ΔOQP
  = 2  x 30  = 60cm²

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°.

Since, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
i.e  ㄥAOB + ㄥCOD = 180^o
⇒  ㄥCOD =180^o  - ㄥAOB
⇒ ㄥCOD =  = 180^o  - 125^o = 55^o

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

(i) (b): OA²=AB²+OB²

\(\Rightarrow \quad O A=\sqrt{10^{2}+5^{2}}=5 \sqrt{5} \mathrm{~cm}\)
(ii) (a) : \(O A=\sqrt{O P^{2}-A P^{2}} \text { (Given) }\)
\(=\sqrt{17^{2}-15^{2}}=\sqrt{64}=8 \mathrm{~cm}\)
(iii) (b): Length of tangent \(A P=\sqrt{O P^{2}-O A^{2}} \text { (Given) }\)
\(=\sqrt{25^{2}-7^{2}}=\sqrt{576}=24 \mathrm{~cm}\)
(iv) (b):

\(\text { Since, } O P=\sqrt{(P T)^{2}+(O T)^{2}}=\sqrt{21^{2}+20^{2}}=29 \mathrm{~cm}\)
(v) (a): Since, OP² + PQ² = OQ²
\(\Rightarrow\) 8² + x² = (x + 2)²\(\Rightarrow\) 64 = 4x + 4\(\Rightarrow\) x = 15 cm
So, length of tangent, PQ = 15 cm.

Here in right angled triangle ABC, AB = 6 m and BC= 8 cm.
\(\therefore\) By Pythagoras theorem \(A C=\sqrt{(A B)^{2}+(B C)^{2}}\)
\(=\sqrt{(8)^{2}+(6)^{2}}=\sqrt{100}=10 \mathrm{~m}\)
Also, AP = x m.
(i) (c):AR=AP = xm      ..(1)
[Since, length of tangents drawn from an external point are equal]
(ii) (b): BQ = BR = AB - AR = (6 - x) m         (Using (1))
(iii) (d): CQ = CP = AC - AP = (10 - x) m
Also, CQ = BC - BQ = BC - BR = 8 - (6 - x) = 2 + x
(iv) (b): Since, CQ = 10 - x = 2 + x
\(\Rightarrow\) 8 = 2x \(\Rightarrow\) x = 4
\(\therefore\)  AR = AP = 4 m, BR = BQ = 2 m
and CP = CQ = 6 m
Also, OQ.\(\perp\)BQ and OR.l BR
\(\therefore\) BROQ is a square.
(v) (a): Radius of the circle, OR = BR = 2 cm

(i) (a)
(ii) (d)
(iii) (b)
(iv) (a)
(v) (c)