(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

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

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

Here, AS = 5 cm, BT = 4 cm  [\(\therefore\)Radii of circles]
(i) (c): Since, radius at point of contact is perpendicular to tangent.
\(\therefore\) By Pythagoras theorem, we have
\(P A=\sqrt{P S^{2}+A S^{2}}=\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{~cm}\)
(ii) (b): Again by Pythagoras theorem, we have
\(B Q=\sqrt{T Q^{2}+B T^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5 \mathrm{~cm}\)
(iii) (d): PK = PA + AK = 13 + 5 = 18 cm
(iv) (c): QY = BQ - BY = 5 - 4 = 1 cm
(v) (c): PS² = PA² - AS² = PA² - AK²
= (PA + AK)(PA - AK) = PK.PX [\(\because\) AK = AX]

(i) (b)
(ii) (c): In \(\Delta\)OAP and \(\Delta\)OBP,
\(\angle\)OAP = \(\angle\)OBP = 90°
[Since, radius at the point of contact is perpendicular to tangent]
OP = OP (Common)
OA = OB (Radii of circle)
So, \(\angle\)OAP == \(\angle\)OBP (By RHS congruency criterion)
(iii) (b): In quadrilateral OAPB, \(\angle\)AOB = 150°  [Given]
\(\angle\)OAP = \(\angle\)OBP = 90°
\(\therefore\) \(\angle\)APB = 360° - 90° - 90° - 150° = 30°
(iv) (d): We have, \(\angle\)APB = 40°

Now,PA =PB [Since, length of tangents drawn  from an external point are equal]
In \(\Delta\)PAB, \(\angle\)PAB = \(\angle\)PBA = 70° [Angles opposite to equal sides are equal]
Also, \(\angle\)PAB + \(\angle\)BAO = 90°
[Since, radius at the point of contact is perpendicular to tangent]
\(\Rightarrow\) \(\angle\)BAO = 90° - 70° = 20°
(v) (d): We have, \(\angle\)ABO = 45°

\(\because\) AO = OB (Radii of circle)
\(\therefore\) \(\angle\)BAO = \(\angle\)ABO = 45°  [Angles opposite to equal sides are equal]
Now, in \(\Delta\)OAB,
\(\angle\)AOB = 180° - 45° - 45° = 90°
Since, \(\angle\)APB = 360° - 90° - 90° - 90° = 90° i.e., AP\(\perp\) BP
So, OAPB is a square.

(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): \(\text { Here, } O A^{2}=O D^{2}+A D^{2}\)
\(\Rightarrow A D=\sqrt{25-9}=4 \mathrm{~cm}\)
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) (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) (b):In \(\Delta\)OAC, \(\angle\)OCA = 90°
[Since, radius at the point of contact is perpendicular to tangent]
\(\therefore\) \(\angle\)OAC + \(\angle\)AOC = 90° \(\Rightarrow\) x + y = 90°
(ii) (c):

Since, OB \(\perp\) PB [Since, radius at the point of contact is perpendicular to tangent]
and \(\angle\)PBA = 50° (Given)
\(\therefore\) \(\angle\)OBA = 90° - 50° = 40°
Also, OA = OB [Radii of circle]
\(\therefore\) \(\angle\)OAB = \(\angle\)OBA = 40° [Angle opposite to equal sides are equal]
(iii) (d): In quadrilateral OQPR,
\(\angle\)ROQ + \(\angle\)RPQ = 180° [\(\because\)Angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the point of contact at the centre]
\(\Rightarrow\) LROQ = 180° - 30° = 150°
(iv) (b): Here, \(\angle\)ABC = 90° (Angle in a semicircle)
So, in \(\Delta\)ABC, \(\angle\)BAC = 180° - 90° - 55° = 35°
Also, \(\angle\)OA T = 90°
\(\Rightarrow\)\(\angle\)BAT + \(\angle\)OAB = 90° \(\Rightarrow\)\(\angle\)BAT = 90° - 35° = 55°
(v) (b): Here, \(\angle\)PAB = 72°
\(\therefore\) \(\angle\)OAB = 90° - 72° = 18°
Also, \(\angle\)AOB = 132° [Given]
Now, in \(\Delta\)OAB, \(\angle\)ABC = 180° - 132° - 18° = 30°

(i) (d):We have, AP = AQ, BP = BD, CQ = CD ...(i)
[\(\because\) Tangents drawn from an external points are equal in length]
Now, AB + BC + AC = 7 + 5 + 8 = 20 cm
\(\Rightarrow\) AB + BD + CD + A C = 20 cm
\(\Rightarrow\)AP + AQ = 20 cm \(\Rightarrow\)2AP = 20 cm \(\Rightarrow\)AP = 10 cm
(ii) (c): Let AF = AE = x cm
[\(\because\) Tangents drawn from an external point to a circle are equal in length]
Given, BD = FB = 9 cm, CD = CE = 3 cm
\(\text { In } \Delta A B C, A B^{2}=A C^{2}+B C^{2} \)
\(\Rightarrow(A F+F B)^{2}=(A E+E C)^{2}+(B D+C D)^{2} \)
\(\Rightarrow(x+9)^{2}=(x+3)^{2}+12^{2} \)
\(\Rightarrow 18 x+81=6 x+9+144 \)
\(\Rightarrow 12 x=72 \Rightarrow x=6 \mathrm{~cm} \)
\(\therefore A B=6+9=15 \mathrm{~cm}\)
(iii) (b): Here, AP = AS = 4 cm
\(\therefore\) DS = DR = 10 - 4 = 6 cm
And BP = BQ = 2 cm. So, CR = CQ = 5 - 2 = 3 cm
So, CD = DR + CR = 6 + 3 = 9 cm
(iv) (d): Here \(\angle\)OAP = 90°
In \(\Delta\)AOP and \(\Delta\)BOP
\(\angle\)OAP = \(\angle\)OBP  [90° each]
OA = OB [Radii of circle]
PA = PB [Tangents drawn from an external point are equal]
\(\therefore\) \(\Delta A O P \cong \Delta B O P\) [By SAS congruency]
\(\therefore\) \(\angle\)APO = \(\angle\)OPB [C.P.C.T]
\(\therefore\) \(\angle\)BPA = 40° + 40° = 80°

(v) (c): For bigger circle, PA = PB ... (i)
[\(\because\)Tangents drawn from an external point are equal in length]
Similarly, for smaller circle, PB = PC ...(ii)
From (i) and (ii), we get
PA = PB = PC = 7 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]
\(\Rightarrow \angle N A S=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{\theta}{2}\)