Class 10 Maths Previous Year Questions - Circles- 1

Ans: (c)
Here, circle with centre O and O' are intersecting at two distinct points A and B. So, in this situation PQ, RS are the tangents which can be drawn.

Ans: (b)
OM ⊥ ML [as tangent from centre is ⊥ at point of contact]

∠OML = 90º 
and ∠NML = 70º 
⇒ ∠OMN = 90º – 70º 
= 20º 
∵ OM = ON = Radii of same circle 
∴ ∠OMN = ∠ONM = 20º
In ∆M ON, 
∠OMN + ∠ONM + ∠MON = 180º 
⇒ 20º + 20º + ∠MON = 180º 
⇒ ∠MON = 140º

Ans: (d)
∠OTP = 90º [Line from centre is ⊥ to tangent at point of contact] 
∠x = ∠TPO + ∠OTP [Exterior Angle Prop.] 
x = 35º + 90º = 125º

Ans: (d)
Since tangent is perpendicular to radius at the point of contact.
∴ ∠PTO = 90°
Hence, by the exterior angle formula, in ΔOTP, we getx = 90° + 25°
= 115°

Ans: (a)
We have ∠AOB = 95°
In ΔAOB, ∠OAB = ∠OBA
Now, ∠OAB + 95° + ∠OBA =180° (Angle sum property of a triangle)


∴ ∠OAB = ∠OBA = 42.5° [From (i)]
Now, OB is perpendicular to the tangent line PQ
∠OBQ = 90°
OA = OB (Radius of circle)
So ∠OAB = ∠OBA
95 + 2x = 180 (Sum of angles of a triangle is 180)
2x = 85
=> x = 42.5
∠ABQ = 90 - 42.5
= 47.5

Ans: (a)
Draw OA ⊥ TA.

In ΔOTA ∠OAT = 90° [∵ Tangent to a circle is perpendicular to the radius passing through the point of contact]
and ∠OTA = 30°

Ans: (b)
Join OR.
We know that tangent to a circle is ⊥ to radius at the point of contact. So, QQ⊥PQ and QR ⊥ PR.
Also, ∠QPR = 90°
Now, in quadrilateral OQPR,
∠QQR - 360o - (90° + 90° + 90°)
= 90°
Also, PQ - PR [∵ Tangents drawn from an external point are equal)
∴ PQQR is a square.
Hence, PQ = OQ = 4 cm

Ans: (a)
OB ⊥ AB     [∵As tangent to a circle is perpendicular to the radius through the point of the contact]
In ΔOAB,

OA² = OB² + AB² [By Pythagoras theorem]
⇒ (41)² = 9² + AB²
⇒ AB² = 41² – 9²
= (41 - 9)(41 + 9)
= (32)(50)
= 1600
⇒ AB =
= 40 cm

Ans: (c)
PR is tangent which touches circle at point P.
So, ∠OPR = 90°
∠OPQ = 90° - ∠RPQ = 90° -  50° =  40°
In, ΔPOQ,
OP = OQ (Radii of circle)
So, ∠OQP = ∠OPQ=40°
⇒ ∠POQ = 180° - 40° - 40° = 100°

Ans: 

OR

Ans: Since, tangents drawn from an external point are equal.
TP = TQ
⇒ ∠TPQ= ∠TQP    _(i)
Angles opposite to equal sides are equal)
In ΔTPQ,
∠PTQ + ∠TQP + ∠TPQ = 180º
⇒ ∠PTQ + ∠TQP + ∠TPQ = 180º    [Using (i)]
⇒ ∠PTQ + 2∠TPQ = 180º
⇒ ∠PTQ = 180º - 2∠TPQ   ...(ii)
Now, OPT = 90° (Tangent is perpendicular  to the radius through the point of contact)
∴ ∠TPQ - 90° - ∠OPQ
From (ii) and (iii), ∠PTQ = 180° - 2(90° - ∠OPQ)
= 180° - 180° + 2∠OPQ = 2∠OPQ

Ans: Given, ∠B = 90°, AD = 17 cm, AB = 20cm, DS = 3 cm
Now, DS = DR and AR = AQ [∵ Tangents drawn from an externa! point to the circle are equal]
∴ DR = 3 cm
AR = AD - DR = 17 - 3 = 14 cm
∴ AQ = 14 cm
Now, BQ = AB - AQ = 20 - 14 = 6 cm
OQ ⊥ BQ, OP ⊥ BP    (∵ Tangent at any point of a circle is perpendicular to the radius through the point of contact)
∴ Quadrilateral BQOP is a square
∴ BQ = OQ = r = 6 cm
Hence, the radius of the circle = 6 cm.

Ans: Let P lie an external point, O be the centre of the circle and PA and PB are two tangents to the circle as shown in  figure.

In ΔQAP and ΔOBP.
OA = OB [Radius of the circle]
OP = OP [common]
PA = PB
[∵ Tangents drawn from an external point to a circle are equal]
So, ΔOAP = ΔOPB
So, ∠APO = ∠BPO
 Hence. OP bisects ∠APB

Ans: Let the centre of the two concentric circlet is O and AB be the chord of the larger circle which touches the smaller circle at point P as shown in figure.
∴ AB is a tangent to the smaller circle at point P
⇒ OP ⊥  AB
By Pythagoras theorem, in ΔOPA

OA² = AP² + OP²
⇒ 5² = AP² +3²
⇒ AP² =5² - 3² = 25 - 9
⇒ AP² = 16 ⇒ AP = 4cm
∴ AB = 2AP = 8cm
In ΔOPB Since, OP ⊥ AB
AP = PB [∵ Perpendicular drawn from the centre of the circle bisects the chord]
∴ AS = 2AP = 2 x 4 = B cm
∴ The length of the chord of the larger circle is a cm.

Ans: Let PA and PB are two tangents on a circle from point P as shown in the figure.
Let is known that tangent to a circle is perpendicular to the radius through the point of contact.
∠OAP =∠OBP = 90°
In quadrilateral AOBP,
∠OAP + ∠APB + ∠PBO + ∠BOA = 360°
90° + ∠APB + 90° + ∠BOA = 360°     [Using (i)]
∠APB + ∠BOA = 360° - 180°
∴ ∠APB + ∠BOA = 180°
Hence proved.

Ans: Since, OP bisects the chord AD, therefore ∠OPA = 90° ....[∵ The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord]
Now, In ΔAOP,
∠A = 180° – 60° – 90°
= 120° – 90°
= 30°
Also, we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact
∴ ∠ABC = 90°
Now, In ΔABC,
∠C = 180° – ∠A – ∠B
= 180° – 30° – 90°
= 150° – 90°
= 60°

Ans: 

Given, ∠ABO = 40°
∠XAO = 90°  ...(Angle between radius and tangent)
OA = OB  ...(Radii of same circle)
⇒ ∠OAB = ∠OBA
∴  ∠OAB = 40°
Now, applying the linear pair of angles property,
we get
∠BAY + ∠OAB + ∠XAO = 180°
⇒ ∠BAY + 40° + 90° = 180°
⇒ ∠BAY + 130° = 180°
⇒ ∠BAY = 180° – 130°
⇒ ∠BAY = 50°
Now, In ΔAOB,
∠AOB + ∠OAB + ∠OBA = 180°
or, ∠AOB + 40° + 40° = 180°
or, ∠AOB = 180° – 80° = 100°
Hence proved.

Ans: Given: Two circles with centres O and O' of radii 2r and r respectively, touch each other internally at A, AB is the chord of bigger circle touches the smaller circle at C.
To prove: C bisects AB i.e. AC = CB
Here, for smaller circle (O' r)
∠ACO = 90°  (Angle in a semicircle is 90°)
∴ OC ⊥ AC
Now, in bigger circle (O, 2r)
Since. AB is a chord and OC ⊥ AB.
AB = CB
[∵ Perpendicular drawn from centre of the circle to a chord bisects the chord]
Hence, C bisects the chord AB.

Ans: It is given that ∠QPR = 90°
We know that the lengths of the tangents drawn from the outer point to the circle are equal.
PQ = PR ... (1)
The radius is Perpendicular to the tangent line at the point of contact.
∴ ∠PQO = 90°
and
∠ORP = 90°
In quadrilateral OQPR:
∠QPR + ∠PQO + ∠QOR + ∠ORP = 360°
⇒ 90° + 90° + ∠QOR + 90° = 360°
⇒ ∠QOR = 360° - 270° = 90°
∴ QPR = ∠PQO = ∠QOR = ∠ORP = 90°
It can be concluded that PQOR is a square.

Ans:  Given, AQ = 5 cm
AQ = AR = 5 cm {v Tangents drawn from an external point to the circle are equal)
Now, AC = AR + RC (∵ OR is a perpendicular bisector of AC AR = RC)
AC = 10 cm

Ans: Given: PQ and PR are tangents from an external point P.
To prove: PQOR is a cyclic quadri lateral.
Proof OR and OQ are tlie radius of circle centred at O, and PR a ltd PQ are tangents.
∠ORP = 90° and ∠OQP = 90°
In quadrilateral PQOR, we have
∠OQP + ∠QOR + ∠ORP + ∠RPQ = 360°
90° + ∠QOR + 90° + ∠RPQ = 360°
180° + ∠QOR + ∠RPQ = 360°
∠QOR + ∠RPQ = 360° - 180°
So, ∠O + ∠P = 180°
∠P and ∠O are opposite angles of quadrilateraI which are supplementary.
∴ PQOR is a cyclic quadrilateral.

Ans: 

Given, radius of circle =5cm
PA and BC are two tangent at point A and B
OP = 13 cm
Step1: OA is perpendicular on tangent AP (OA is radius of the circle)
In right angle triangle AOAP
(OP)² = (OA)² + (AP)²
⇒ (AP)² = (OP)² – (0A)²
⇒ (AP)² = (13)² - (5)² = 169 - 25 = 144
AP = √144 = 12
AP = PA = 12 cm
Step 2: Let length of BC be x
But AC = BC= x (tangent from an external point)
So length of PC = 12 - x and PB = OP - OB = 13 - 58cm
(OB is the radius and length of OP is given)
OB is perpendicular on tangent CB , so ∠OBC = ∠CBP = 90 °
In right angle triangle ΔCBP
(CP)² = (BP)² + (BC)²
⇒ (12 - x)² = (8)² + (x)²
⇒ 144 - 2x + x² = 64 + x²
⇒ 144 - 24x - 64 = 0
⇒ 80 - 24x = 0 ⇒ x = 80/24 = 3.33cm
Hence, the length of BC is 3.33 cm and PA is 12 cm

Ans:  In the given figure,
PQ = 8 cm and OP = 5 cm

OR ⊥ PQ and so, OR bisects PQ. [ ∵ Perpendicular drawn from the center to the chord bisects the chord]
⇒ PR = RQ = 4 cm
In Δ POR ,
OP² = OR² + PR²
⇒ 5² = OR² + 4²
⇒ OR = 3 cm
In ΔTPO and ΔPRO,
∠ TOP = ∠ ROP [common]
and ∠ TPO = ∠ PRO [each 90º]
∴ Δ TPO and Δ PRO are similar. [by AAA Similarity]

[∵Tangents drawn from an external point to a circle are equal in length]

Ans: Given : A parallelogram ABCD circumscribing a circle with centre O.
To prove : ABCD is a rhombus.
Proof: We know that the tangents drawn to a circle from an external Doint are eaual in length.

⇒ AP = AS  [Tangents drawn from A]    ...(i)
⇒ BP = BQ    [Tangents drawn from B]    ...(ii)
⇒ CR= CQ    [Tangents drawn from C]    ...(iii)
⇒ DR = DS    [Tangents drawn from D]    ...(iv)
Adding (i), (ii), (iii) and (iv) we get
AP + BP +  CR + DR = AS + BQ + CQ + DS
= (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ 2AB = 2BC         [Opposite sides of the given parallelogram are equal ∴ AB = DC and AD = BC)
AB = BC = DC = AD
Hence, ABCD is a rhombus.

Ans: Given: A circle is touching a side QR of ΔPQR at point S.
PQ and PR are produced at M and N respectively.
To prove:
Proof: PM = PN  ...(i) (Tangents drawn from an external point P to a circle are equal)
QM = QS  ...(ii) (Tangents drawn from an external point Q to a circle are equal)
RS = RN  ...(iii) (Tangents drawn from an external point R to a circle are equal)
Now, 2PM = PM + PM
= PM + PN  ...[From equation (i)]
= (PQ + QM) + (PR + RN)
= PQ + QS + PR + RS  ...[From equations (i) and (ii)]
= PQ + (QS + SR) + PR
= PQ + QR + PR

Hence proved.

Ans: 

According to the question,
In a right angle ΔABC is which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P.
Also PQ is a tangent at P
To Prove: PQ bisects BC i.e. BQ = QC
Proof: ∠APB = 90°  ...[Angle in a semicircle is a right-angle]
∠BPC = 90° ...[Linear Pair]
∠3 + ∠4 = 90° ...[1]
Now, ∠ABC = 90°
So in ΔABC
∠ABC + ∠BAC + ∠ACB = 180°
90° + ∠1 + ∠5 = 180°
∠1 + ∠5 = 90°  ...[2]
Now, ∠1 = ∠3  ...[Angle between tangent and the chord equals angle made by the chord in alternate segment]
Using this in [2] we have
∠3 + ∠5 = 90°  ...[3]
From [1] and [3] we have
∠3 + ∠4 = ∠3 + ∠5
∠4 = ∠5
QC = PQ  ...[Sides opposite to equal angles are equal]
But also, PQ = BQ  ...[Tangents drawn from an external point to a circle are equal]
So, BQ = QC
i.e. PQ bisects BC.

Ans: Let common tangent at P meets the tangent AB at C. Since, tangents drawn from an external point to a circle are  equal

∴ AC = CP
and BC = CP
⇒ ∠CAP = ∠CPA = x (say) ...(i)
and ∠CBP = ∠CPB = y (say) ...(ii)
Now, ∠ACP+ ∠BCP = 180°  [Linear pair] ...(*)
In ΔACP, ∠ACP + ∠CPA + ∠CAP = 180° ...(iii)
and in ΔBCP, ∠BCP+ ∠CPB + ∠CBP = 180°...(iv)
Adding (iii) and (iv), we get
∠ACP + x + x + ∠BCP + y + y = 360°
∠ACP + ∠BCP + 2x + 2y = 360°   [Using (i) & (ii)]
= 2(x + y) = 360° - 180° = 180°[Using ('))
⇒ x + y = 90°
i.e., ∠CPA + ∠CPB = 90° => ∠APB = 90°

Ans: Given TP and SP are tangents from an external point P.
PT = PS = 4 cm (v Tangents drawn from an external point to the circle are equal)
∠PTS = ∠PST
(∵ Angles opposite to equal sides are equal) In A TPS, by angle sum property
∠TPS = ∠PTS =∠PST = 60°
⇒ ΔTPS is an equilateral triangle.

∴ TP = PS = TS = 4 cm
∠OSP = 90º and ∠TSP = 60º
∴ ∠OSD = 30º

Ans: (a)
Given that
∠ AOB = 100°
Since OA = OB
So  ∠ OAB =  ∠ OBA = 40°
Since PQ is tangent on the circle. So OB is perpendicular to PQ.
So,
∠ OBP = 90°
∠ OBA + ∠ ABP = 90°
∠ ABP  = 90 – ∠ OBA
∴ ∠ ABP  = 90° – 40°
∴ ∠ ABP  = 50°

Ans: (d)
Since ∠TPO = 25° and ∠OTP  = 90°
x = ∠OTP + ∠TPO
= 90° + 25° = 115°
[∵ Radius is perpendicular to the tangent T]

Ans: (b)
It is known that the length of the tangents drawn from an external point to a circle are equal.
∴ QP = PT= 3.8 cm and PR = PT = 3.8 cm
Now, QR = QP + PR = 3.8cm + 3.8cm = 7.6 cm

Ans: (a)
In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB +∠OAP = 360°
⇒ ∠AOB + 90° + 90° + 80° = 360°
⇒ ∠AOB = 360° - 260° = 100°

Ans: Let the circle touches the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively Since, lengths of tangents drawn from an external point to the circle are equal.

AP = AS    ...(1)    (Tangents drawn from A)
BP = BQ    ...(2)    (Tangents drawn from B)
CR = CQ    ...(3)    (Tangents drawn from C)
DR = DS    ...(4)    (Tangents drawn from D)
Adding (1), (2), (3) and (4), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + PB) + (CR + RD) = (AS + SD) + (BQ + QC)
⇒ AB + CD = AD + BC

Ans: As we know that tangents drawn from an external point are equal in length.
∴ BP = BD and CD = CQ  -----(i)
Also, AP = AQ = 12 cm
⇒ AB + BP = 12 cm and AC + CQ = 12 cm
⇒ AB + BD = 12 cm
AC + CD = 12 cm ...(iii) [Using (i)]
Now, perimeter of ΔABC = AB + BC + CA
= AB + BD + DC + AC
= 12 + 12   [Using (ii)]
= 24 cm

Ans: Since, tangents drawn from an external point are equal
TP = TQ
⇒ ∠TPQ = ∠TQP    ...(i) (∵ Angles opposite to equal sides are equal)
In ΔTPQ,
∠PTQ + ∠TQP + ∠TPQ = 180°
⇒ ∠PTQ + ∠TPQ + ∠TPQ = 180°    [Using (i)]
⇒ ∠PTQ + 2 ∠TPQ = 100°
⇒ ∠PTQ= 180° - 2 ∠TPQ    ...(ii)
Now, ∠OPT = 90° (∵ Tangent is perpendicular to the radius through the point of contact) ZTPQ = 90° - ZOPQ    -Oil)
From (ii) and (iii), ∠PTQ = 180° - 2(90° - ∠OPQ)
= 180° - 180° + 2 ∠OPQ = 2 ∠OPQ

Ans:  Let BL = x, BN = x

[∵ Tangents drawn from an external point to to the circle are equal in length]
CL = CM = 8 - x    [∵ BC = 8cm]
AN = AM = 10 - x    [∵ AB = 10cm]
But AC= 12cm[Given]
∴ AM + MC = 12
10 - x + 8 - x = 12
⇒ 18 - 2x = 12
⇒ 6 = 2x
⇒  x = 3
Length of BL = 3cm
Length of CM = 8 - 3 = 5 cm
Length of AN = 10 - 3 = 7 cm

Ans: 

Given : A circle C(O, r)with diameter AB and let PQ and RS be the tangents drawn to the circle at point A and B.
To prove: PQ || RS
Proof: Since tangent at a point to a circle is perpendicular to the radius through the point of contact.
∴ AB ⊥ PQ and AB ⊥ R S
⇒ ∠PAB = 90° and ∠ABS = 90°
⇒ ∠PAB = ∠ABS
⇒ PQ || RS  [∵ ∠PAB and ∠ABS are alternate interior angles]

Ans: Given: r₂ - r₁ = 7 (r₂ > r₁) ...(i)

(From equation (i))
..... (ii)
Adding (i) and (ii), we get
2r₂ = 56
⇒ r₂ = 28 cm
Also, r₁ = 21 cm (From equation (i))
∴ Radius of simaller circle = 21 cm.

Ans: Given: A line T tangent to the circle at point T and O is the centre of circle.
To prove: OT ⊥ l
Construction: Take point T₁,T₂ and T₃ on line l and join OT₁, OT₂, OT₃
Proof: We observe that points T₁, T₂, T₃ lie outside the circle, whereas point T lies on the circle.
Hence OT₁ > OT
OT₂ > OT
OT₃ > OT
All distances OT₁, OT₂, OT₃ are greater than OT.
Only OT is the shortest distance.
Also OT = r
Hence r is the shortest distance from tangent l of the circle to the centre as we know that shortest distance between the point on line is perpendicular distance
So, OT ⊥ l

Ans: Given: PQ is a chord of length 8 cm
Radius OP = 5 cm. PT and QT are tangents to the circle.
To find: TP
OT is perpendicular bisector of PQ
∠ORP = 90°
(Line joining the centre of circle to the common point of two tangents drawn to circle is perpendicular bisector of line joining the point of contact of the tangents.


⇒

x² = 16
⇒ x = 3


(OR + RT)² = 25 + PT²
(3 + y)² = 25 + PT²
9 + y² + 6y = 25 + PT² .......(i)
In ΔPRT, TP² = PR² + RT²
⇒ PT² = (4)² + (y)² .......(ii)
Put value of PT² in eq (i)
9 + y² + 6y = 25 + 16 + y²
6y = 25 + 16 - 9
6y = 32
y = 32/6 = 16/3 cm
Putting y = 16/3 cm in eq (ii), we get

Ans: Join OA and OB.
In ΔOAP and ΔOBP.

OA = OB [Radii of same circle]
PA = PB [Tangents from external point]
OP = OP [Common]
So ΔOAP ≅ ΔOBP (By SSS rule)
∠1 = ∠2 [By C.P.C.T.]
Now, In ΔATP and ΔBTP.

 PA = PB [Tangents from external point]

 AT = BT [By C.P.C.T.] ...(1)

 ∠ATP = ∠BTP [By CPCT] ...(2)
Since, ATB is a straight line. 
∠ATP + ∠BTP = 180º 
⇒ ∠ATP + ∠ATP = 180º [From (2) ] 
⇒ 2∠ATP = 180º 
⇒ ∠ATP = 90º ...(3) 
From (1) and (3) we can say that OP is ⊥ bisector of AB

Ans: Given, PQ is a tangent to a circle, and QOR is a diameter of a circle. 
Also, ∠POR = 130º 
Now, ∠RST = 1/ 2 ∠TOR
∠2 = 1/ 2 × 130º = 65º ...(i) [∠TOR = ∠POR] 

[Since, angle subtended by the arc at centre is twice the angle subtended by it on any remaining part of the circle]
Since, ROQ is the diameter of the circle 
∴ ∠ROT + ∠QOT = 180º 
∠QOT = 180º – 130º = 50º ...(ii) 
and ∠PQR = 90º ...(iii) [tangent at any point of a circle is perpendicular to the radius through the point of contact]
In ΔQOP 
∠QOP + ∠PQO + ∠OPQ = 180º 
⇒ 50º + 90º + ∠1 = 180º [from (ii) and (iii)] 
∠1 = 180º – 140º = 40º ...(iv)
∴ ∠1 + ∠2 = 40º + 65º [From (i) and (iv)] 
= 105º 
Hence, the sum of ∠1 + ∠2 is 105º

Ans: Given: area (∆ABC) = 84 cm²
Radius of circle, r = OP = OQ = OR = 4 cm AP = 6 cm and BP = 8 cm 
Now AP = AR = 6 cm [Q two tangents from an external point to a circle are equal] 
Similarly, BP = BQ = 8 cm 
and QC = RC = x (say) 
AC = 6 + x 
and BC = 8 + x 
Now, area (∆ABC) = area (∆AOB) + area (∆BOC) + area (∆AOC)

⇒ 84 = 28 + (16 + 2x) + (12 + 2x)
⇒ 84 = 56 + 4x 
⇒ 4x = 84 – 56 
⇒ 4x = 28 
⇒ x = 7 
Hence, AC = 6 + 7 = 13 cm 
and BC = 8 + 7 = 15 cm.

Ans: Given, ∠QPR = 120°
Radius is perpendicular to the tangent at the point of contact.
∠OQP = 90° ⇒ ∠QPO = 60°

(Tangents drawn to a circle from an external point are equally inclined to the segment, joining the centre to that point.)

Ans: PA = PB (Tangents from an external point are equal)
and ∠APB = 60°
⇒ ∠PAB = ∠PBA = 60°
∴ ΔPAB is an equilateral triangle.
Hence AB = PA = 5 cm.

Ans: Let ∠TOP = θ
∴
Hence, ∠TOS = 120°
In ∠OTS, OT = OS    (Radii of circle)
⇒

Ans: OA = 6 cm, OB = 4 cm, AP = 8 cm
OP² = OA² + AP² = 36 + 64 = 100
⇒ OP = 10 cm
BP² = OP² - OB² = 100 - 16 = 84
⇒

Ans: ∵ PA = PB ⇒ ∠BAP = ∠ABP = 50°
∴  ∠APB = 180° - 50° - 50° = 80°
and ∠AOB = 180° - 80° = 100°

Ans: ∠ACB = 90° (Angle in the semicircle)
∠CAB = 30° (given)
In ΔABC,
90° + 30° + ∠ABC = 180°
⇒ ΔABC = 60°
Now, ∠PCA = ∠ABC (Angles in the alternate segment)
∴  ∠PCA = 60°

OR
Construction: Jo in 0 to C.
∠PCO = 90° (∵ Line joining centre to point of contact is perpendicular to PQ)
In ΔAOC, OA = OC (Radii of circle)
∴ ∠OAC = ∠OCA = 30° (Equal sides have equal opp. angles)
Now, ∠PCA = ∠PCO - ∠ACO
= 90° - 30° = 60°

Ans:  Given. A circle C(0, r). PA and PB are tangents to the circle from point P, outside the circle such that ∠APB = 120°. OP is joined.

To Prove. OP = 2AP.
Construction. Join OA and OB.
Proof. Consider Δs PAO and PBO
PA = PB [Tangents to a circle, from a point outside it, are equal.]
OP = OP [Common]
∠OAP = ∠OBP = 90°

Ans: AC is tangent to circle with centre O.
Thus ∠ACO = 90°
In ΔAO'D and ΔAOC
∠ADO' = ∠ACO = 90º
∠A = ∠A  (Common)

Ans: 

(Tangents from an external point to a circle are equal)
In right ΔAET.
TA² = TE² + EA²
⇒ (12 - x)² = 64 + x²    ⇒    144 + x² - 24x = 64 + x²
⇒  x = 80/24    ⇒ x = 3.3 cm
Thus, AB = 6.6 cm

Ans: Given, ∠QPT = 60°
So, OP is the radius of the circle. 
Now, ∠OPT = 90° 
∠OPQ = ∠OPT – ∠QPT = 90° – 60° = 30° 
In ΔOPQ, OP = OQ [radii of circle] 
∠OQP = ∠OPQ = 30° [Q Angles opposite to equal sides are equal] 
∠POQ = 180° – (30° + 30°) = 120° 
Reflex ∠POQ = 360° – 120° = 240° 
We know that, angle subtended by an arc at the centres double the angle subtended by it on the remaining part of the circle. 
so, ∠PRQ = 1/ 2 Reflex ∠POQ
Therefore, 240 / 2 ° = 120°

Ans: Let ABCD be a parallelogram such that its sides touch a circle with centre O.
We know that the tangents to a circle from an exterior point are equal in length.
Therefore, we have
AP = AS             (Tangents from A)      ... (i)
BP = BQ             (Tangents from B)     ... (ii)
CR = CQ              (Tangents from C)     ... (iii)
And DR = DS      (Tangents from D) ... (iv)

Adding (i), (ii), (iii) and (iv), we have
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = BC + BC      (∵ ABCD is a parallelogram ∴ AB = CD, BC = DA)
2AB = 2BC ⇒ AB = BC
Thus, AB = BC = CD = AD
Hence, ABCD is a rhombus.

Ans: Given: AP and AQ are two tangents from a pointed to a circle C (O, r).
To Prove: AP = AQ
Construction: Join OP, OQ and OA.
Proof: In order to prove that AP = AQ, we shall first prove that ΔOPA ≅ ΔOQA.
Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OP ⊥ AP and OQ ⊥ AQ
⇒ ∠OPA = ∠OQA = 90°    ........(i)

Now, in right triangles OPA and OQA, we have
OP = OQ    (Radii of a circle)
∠OPA = ∠OQA    (Each 90°)
and OA = OA    (Common)
So, by RHS-criterion of congruence, we get

Hence, lengths of two tangents from an external point are equal.