\(\because \)  The tangents drawn from an external point to the circle are equal
\(\therefore \) CR = CP = 6 cm
BQ = BP = 10 cvm
AR = AQ = AB - BQ
= 18 cm - 10 cm = 8 cm
\(\therefore \)  x = AC = AR + CR = 89 cm +  6 cm = 14 cm

5 cm

Given: BD is a diameter of the circle with centre O, ABCD is a cyclic quadrilateral. 

o find: \(\angle \)BCP 
Sol. Since BD is the diameter of the circle. 
⇒ \(\widehat { BCD } \) is a semicircle. 
⇒ BCD = 90^o [Angle in a semicircle] 
But, \(\angle \)BCP + \(\angle \)BCD + \(\angle \)DCQ = 180^o 
[Angles on the same point of a straight line on the same side] 
⇒  \(\angle \)BCP + 90^o + 40^o = 180^o 
⇒ \(\angle \)BCP = 180^o - 130^o = 50^o

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.)

∠LOM= 90^o

Given \(\triangle ABC\)in which AB = AC.
To prove BC = 2 CE
Proof: We know that, tangents from an external point to a circle are equal in length.
\(\therefore\)  AD = AF   [A is an external point] ...........(i)
BD = BE   [B is an external point] ...........(ii)
and   CE = CF   [C is an external point] ............(iii)
Now, AB = AC  [given]
\(\Rightarrow\)   AB - AD = AC - AD     [subtracting AD from both sides]
\(\Rightarrow\)   AB - AD = AC - AF    [using Eq. (i)]
\(\Rightarrow\)   BD = CF
\(\Rightarrow\)  BE = CF      [using Eq. (ii)]
\(\Rightarrow\)   BE = CE     [using Eq. (iii)] ........(iv)
Now, BC = CE+BE = CE+CE  [from Eq. (iv)]
 \(\Rightarrow\)   BC = 2CE                                      
Hence proved.

Given: Let PQ and GH are the common tangents to both circles and common tangent at C meets PQ at E and GH at F. 
To Prove: EF bisects PQ and GH. 
Proof: Tangents'drawn from an external point to a circle are equal. 
EP = EC and EQ = EC
⇒ EP = EQ
⇒ PQ is bisected by EF at E. 
Similarly, GH is bisected by EF at F. 
ஃ The common tangent EF drawn at 'C' bisects the other two common tangents. 

Given: The two concentric circles with centre O. AB is the chord of larger circle which touches the smaller circle at C. 
To Prove: AC = CB 
Construction: Join OA, OC and OB. 
Proof: AB is a tangent to inner circle at point C. OC is the radius drawn at the point Of contact. 
OC ⊥ AB
⇒ \(\angle \)1 = \(\angle \)2 = 90°
In rt. \(\triangle\)AOC and \(\triangle\)BOC 
Hypotenuse OA = Hypotenuse 0B     [radii Of larger circle] 
OC = OC     [common]
\(\triangle\)AOC = \(\triangle\)BOC [by RHS congruency axiom] 
AC = CB   [c.p.c.t.]

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

(a) (iv) NS and NR are both tangent to the circle
So, \(N S \perp S A\) and \(N R \perp R A\) 
[ \(\therefore\)  Tangent to a circle is perpendicular to the radius through the point of contact ]
\(\therefore \angle N S A=90^{\circ}\) 
(b) (ii) 
\(\angle N S A=90^{\circ} \quad \Rightarrow N A^{2}=N S^{2}+S A^{2}\) [ By Pythagora's Theorem]
\(\Rightarrow N S=\sqrt{N A^{2}-S A^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{169-25}\) 
\(=\sqrt{144}=12 \mathrm{~m}\) 
(c) (iiii)
NR = NS = 12m
Tangents drawn from an external point are equal
\(\therefore N R=12 \mathrm{~m}\) 
(d) (iv)
\(\angle S N R+\angle R A S=180^{\circ} \Rightarrow \angle R A S=180^{\circ}-\angle S N R\) 
[\(\because\) The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact to the centre]
\(\Rightarrow \angle R A S=180^{\circ}-\theta\) 
(e) (i) 
\(\angle R A S=180^{\circ}-\theta\) [ Calulated bove]
Now, \(\angle N A S=\angle N A R=\frac{1}{2} \angle R A S\) 
[\(\because\) If two tangents are drawn from an external point, then they extend equal angles at the centre]
\(\Rightarrow \angle N A S=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{\theta}{2}\)

(i) (d):

Two tangents of a circle are parallel only when they are drawn at ends of a diameter.
So,PQ is the diameter of the circle.
(ii) (b)
(iii) (d)
(iv) (a): Here, the two circles have a common point of contact T and PQ is the tangent at T. So, PQ is the tangent to both the circles.
(v) (a)

(i) (a): \(\text { Here, } O A^{2}=O D^{2}+A D^{2}\)
\(\Rightarrow A D=\sqrt{25-9}=4 \mathrm{~cm}\)
As OD bisects AB, then
AB = 2AD = 2 x 4 = 8 cm

(ii) (c): Here, PB² + OB² = OP² = PA² + OA²
Then PB² + 9 = 144 + 25\(\Rightarrow\) PB2 = 160
\(\Rightarrow\) P B = 4\(\sqrt{10}\) cm
(iii) (b): Here, OP² - PB² = OB² and OP² - PA² = OA²
\(\therefore O B=\sqrt{100-64}=\sqrt{36}=6 \mathrm{~cm}\)
\(\text { and } O A=\sqrt{100-36}=\sqrt{64}=8 \mathrm{~cm} \)
\(\therefore \quad A B=O A-O B=8-6=2 \mathrm{~cm}\)
(iv) (d):Here, in right angled \(\Delta\)OBD, OB = 5 cm and OD=3 cm.
\(\therefore B D=\sqrt{25-9}=\sqrt{16}=4 \mathrm{~cm}\)
Since, chord BP is bisected by radius OD.
\(\therefore\) BP = 2BD = 8 cm
(v) (a):Let x be the radii of smaller circle.

\(\text { Now, } O A^{2}=O D^{2}+A D^{2} \)
\(\Rightarrow(x+4)^{2}=x^{2}+12^{2} \)
\(\Rightarrow 8 x+16=144 \)
\(\Rightarrow x=16 \mathrm{~cm}\)

(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.

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

(i) (b): 

Here, OS the is radius of circle.
Since radius at the point of contact is perpendicularto tangent.
So, \(\angle\)OSA = 90°
(ii) (d): Since, length of tangents drawn from an external point to a circle are equal.
\(\therefore\) AS=AP,BP=BQ,
CQ = CR and DR = DS
(iii) (a): AP = AS = AD _ DS = AD _ DR (Using (1)
= 11 - 7 = 4 cm
(iv) (b): In quadrilateral OQCR,

\(\angle\)QCR = 60° (Given)
And \(\angle\)OQC = \(\angle\)ORC = 90° [Since, radius at the point of contact is perpendicular to tangent.]
\(\therefore\) \(\angle\)QOR = 360° - 90° - 90° - 60° = 120°
(v) (c): From (1), we have AS = AP, DS = DR,
BQ = BP and CQ = CR
Adding all above equations, we get
AS + DS + BQ + CQ = AP + DR + BP + CR
\(\Rightarrow\) AD + BC = AB + CD