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 ∆MON, 
∠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 get x = 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: 

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ TP = TQ ... (1)

∴ ∠TQP = ∠TPQ (angles of equal sides are equal) ... (2)

Now, PT is tangent, and OP is the radius.

∴ OP ⊥ TP (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

∴ ∠OPT = 90°

or, ∠OPQ + ∠TPQ = 90°

or, ∠TPQ = 90° - ∠OPQ ... (3)

In ∆TPQ,

∠TPQ + ∠PQT + ∠QTP = 180° (Sum of angles of a triangle is 180°)

or, 90° - ∠OPQ + ∠TPQ + ∠QTP = 180°

or, ∠(90° - ∠OPQ) + ∠TPQ + ∠QTP = 180° [from (2) and (3)]

or, 180° - 2∠OPQ + ∠TPQ = 180°

or, 2∠OPQ = ∠TPQ - proved

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 = 8 cm
∴ The length of the chord of the larger circle is 8 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)

A tangent at any point of a circle is perpendicular to the radius at the point of contact.

In ΔOAP and in ΔOBP:

• OA = OB (radii of the circle are always equal)
• AP = BP (length of the tangents)
• OP = OP (common)

Therefore, by SSS congruency ΔOAP ≅ ΔOBP.

SSS congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

If two triangles are congruent, then their corresponding parts are equal.

Hence:

• ∠POA = ∠POB
• ∠OPA = ∠OPB

Therefore, OP is the angle bisector of ∠APB and ∠AOB.

Hence, ∠OPA = ∠OPB = 1/2 (∠APB)
= 1/2 × 80°
= 40°

By the angle sum property of a triangle, in ΔOAP:
∠A + ∠POA + ∠OPA = 180°

OA ⊥ AP (Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.)

Therefore, ∠A = 90°

90° + ∠POA + 40° = 180°
130° + ∠POA = 180°
∠POA = 180° - 130°
∠POA = 50°

Thus, option (A) 50° is the correct answer.

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: 

Step 1: Identify the Tangents
From the problem, we know that AP and AQ are tangents to the circle from point A, and BC is also a tangent. According to the properties of tangents from an external point, the lengths of the tangents drawn from the same external point to a circle are equal.

Step 2: Set Up the Equations
Since AP = 12 cm, we can conclude that:

• AP = AQ = 12 cm (Equation 1)

Step 3: Identify Other Tangents
From point B, the tangents BD and BP are equal:

• BD = BP (Equation 2)

From point C, the tangents CD and CQ are equal:

• CD = CQ (Equation 3)

Step 4: Express Perimeter of Triangle ABC
The perimeter of triangle ABC can be expressed as:
Perimeter = AB + BC + AC

Step 5: Substitute for BC
Since BC is composed of the tangents from B and C:
BC = BD + CD

Thus, we can rewrite the perimeter as:
Perimeter = AB + (BD + CD) + AC

Step 6: Express AB and AC in Terms of Tangents
From the properties of tangents:

• AP = AB + BP
• AQ = AC + CQ

Substituting BP and CQ with BD and CD respectively, we have:

• AP = AB + BD (Equation 4)
  AQ = AC + CD (Equation 5)

Step 8: Simplify the Expression
Since AP = AQ and both are equal to 12 cm:
Perimeter = (12 - BD) + (BD + CD) + (12 - CD)

This simplifies to:
Perimeter = 12 + 12 = 24 cm

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: 

Referring to the figure:

OA = OC (Radii of circle)

Now OB = OC + BC

∴ OB > OC (OC being radius and B any point on tangent)

⇒ OA < OB

B is an arbitrary point on the tangent.

Thus, OA is shorter than any other line segment joining O to any point on the tangent.

Shortest distance of a point from a given line is the perpendicular distance from that line.

Hence, the tangent at any point of the circle is perpendicular to the radius.

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.