Q1: The maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is: (2024)
(a) 4
(b) 3
(c) 2
(d) 1
Ans: (c)
Q2: In the given figure, O is the centre of the circle. MN is the chord and the tangent ML at point M makes an angle of 70° with MN. The measure of ∠MON is: (2024)
(a) 120º
(b) 140º
(c) 70º
(d) 90º
Ans: (b)
Q3: In the given figure, if PT is a tangent to a circle with centre O and ∠TPO = 35º, then the measure of ∠x is (2024)
(a) 110º
(b) 115º
(c) 120º
(d) 125º
Ans: (d)
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
Q5: In the given figure. TA is a tangent to the circle with centre O such that OT = 4 cm, ∠OTA= 30^{o}, then length of TA is (2023)
(a) 2√3 cm
(b) 2cm
(c) 2√2 cm
(d) √3 cm
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°
Q6: In figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. If ∠QPR= 90°. then length of PQ is (2023)
(a) 3 cm
(b) 4 cm
(c) 2 cm
(d) 2√2 cm
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
Q7: The length of tangent drawn to a circle of radius 9 cm from a point 41 cm from the centre is (2023)
(a) 40 cm
(b) 9 cm
(c) 41 cm
(d) 50 cm
Ans: (a)
OB ⊥ AB [∵As tangent to a circle is perpendicular to the radius through the point of the contact]
In ΔOAB,
OA^{2} = OB^{2} + AB^{2} [By Pythagoras theorem]
⇒ (41)^{2} = 9^{2} + AB^{2}
⇒ AB^{2} = 41^{2} – 9^{2}
= (41  9)(41 + 9)
= (32)(50)
= 1600
⇒ AB =
= 40 cm
Q8: In the given figure. O is the centre of the circle and PQ is the chord. If the tangent PR at P makes an angle of 50° with PQ, then the measure of ∠POQ is (2023)
(a) 50°
(b) 40°
(c) 100°
(d) 130°
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°
Q9: Case Study: The discus throw is an event in which an athlete attempts to throw a discus. The athlete spins anticlockwise around one and a half times through a circle, then releases the throw. When released, the discus travels along tangent to the circular spin orbit.
In the given figure, AB is one such tangent to a circle of radius 75cm. Point O is the centre of the circle and ∠ABO = 30°. PQ is parallel to OA.
Based on above information
(a) Find the length of AB.
(b) Find the length of OB.
(c) Find the length of AP.
OR
Find the value of PQ. (2023)
Ans:
OR
Q10: Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. (2023)
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
Q11: In the given figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 17 cm, AB = 20 cm and DS = 3 cm, then find the radius of the circle. (2023)
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.
Q12: From an external point, two tangents are drawn to a circle. Prove that the line joining the external point to the centre of the circle bisects the angle between the two tangents. (2023)
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
Q13: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. (2023)
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
OA^{2} = AP^{2} + OP^{2}
⇒ 5^{2} = AP^{2} +3^{2}
⇒ AP^{2} =5^{2}  3^{2} = 25  9
⇒ AP^{2} = 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.
Q14: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the linesegment joining the points of contact at the centre. (2023)
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°
Q16: In Fig. XAY is a tangent to the circle centred at 0. If ∠ABO = 40°. Then find ∠BAY and ∠AOB (2022)
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.
Q17: In Figure, two circles with centres at O and O' of radii 2r and r, respectively, touch each other internally at A. A chord AB of the bigger circle meets the smaller circle at C. Show that C bisects AB. (2022)
View Answer
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.
Q18: In Figure, PQ and PR are tangents to the circle centred at O. If ∠OPR = 45°, then prove that ORPQ is a square. (2022)
View Answer
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.
Q19: In Fig., there are two concentric circles with centre O. If ARC and AQB are tangents to the smaller circle from point A lying on the larger circle, find the length of AC if AO = 5 cm. (2022)
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
Q20: In Figure, O is the centre of the circle. PQ and PR are tangent segments. Show that the quadrilateral PQOR is cyclic. (2022)
View Answer
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.
Q21: In Figure O is centre of a circle of radius 5 cm. PA and BC are tangents to the circle at A and B respectively. If OP = 13 cm. then find the length of tangents PA and BC. (2022)
View Answer
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 + x^{2 }= 64 + x^{2}
⇒ 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
Q22: In fig. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q meet at a point T. Find the length of TP. (2022)
View Answer
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 ,
OP^{2} = OR^{2} + PR^{2}
⇒ 5^{2} = OR^{2} + 4^{2}
⇒ 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]
Q23: Prove that a parallelogram circumscribing a circle is a rhombus. (2022)
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.
Q24: In fig, if a circle touches the side QR of ΔPQR at S and extended sides PQ and PR at M and N, respectively, then
Prove that (2022)
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.
Q25: In figure, a triangle ABC with ∠B = 90° is shown. Taking AB as diameter, a circle has been drawn intersecting AC at point P. Prove that the tangent drawn at point P bisects BC. (2022)
View Answer
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 rightangle]
∠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.
Q26: In figure, two circles touch externally at P. A common tangent touches them at A and B, and another common tangent is at P, which meets the common tangent AB at C. Prove that ∠APB = 90°. (2022)
View Answer
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°
Q27: In the given figure, PT and PS are tangents to a circle with centre O, from a point P such that PT = 4 cm and ∠TPS = 60°. Find the length of the chord TS. Also, find the radius of the circle. (2021)
View Answer
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°
Q29: In the given figure, PT is a tangent at T to the circle with centre O. If ∠TPO = 25°, then x is equal to (2020)
(a) 25°
(b) 65°
(c) 90°
(d) 115°
Ans: (d)
Since ∠TPO = 25° and ∠OTP = 90°
x = ∠OTP + ∠TPO
= 90° + 25° = 115°
[∵ Radius is perpendicular to the tangent T]
Q30: In the given figure, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P If PT = 3.8 cm, then the length of QR(in cm] is (2020)
(a) 3.8
(b) 7.6
(c) 5.7
(d) 1.9
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
Q31: In Figure, if tangents PA and PB from an external paint P to a circle with centre O are inclined to each other at an angle of 80° then ∠AOB is equal to (2020)
(a) 100°
(b) 60°
(c) 80°
(d) 50°
Ans: (a)
In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB +∠OAP = 360°
⇒ ∠AOB + 90° + 90° + 80° = 360°
⇒ ∠AOB = 360°  260° = 100°
Q32: In figure, a quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = BC + AD. (2020)
View 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
Q33: In figure, find the perimeter of ΔABC if AP =12 cm. (2020)
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
Q34: In the figure, two tangents, TP and TQ, are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. (2020)
View Answer
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
Q35: In the given figure, a circle is inscribed in a ΔABC having sides BC = 6 cm, AB = 10 cm and AC = 12 cm. Find the lengths BL, CM and AN. (2019)
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
Q36: Prove that tangents drawn at the ends of a diameter of a circle are parallel. (2019)
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: r_{2}  r_{1} = 7 (r_{2} > r_{1}) ...(i)
(From equation (i))
..... (ii)
Adding (i) and (ii), we get
2r_{2} = 56
⇒ r_{2} = 28 cm
Also, r_{1} = 21 cm (From equation (i))
∴ Radius of simaller circle = 21 cm.
Q38: Prove that the tangent drawn at any point of a circle is perpendicular to the radius through the point of contact. [AI (C) 2017]
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 T_{1},T_{2} and T_{3} on line l and join OT_{1}, OT_{2}, OT_{3}
Proof: We observe that points T_{1}, T_{2}, T_{3} lie outside the circle, whereas point T lies on the circle.
Hence OT_{1} > OT
OT_{2} > OT
OT_{3} > OT
All distances OT_{1}, OT_{2}, OT_{3} 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
Q39: In Fig., PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents drawn at P and Q intersect at T. Find the length of TP. [AI (C) 2017, Foreign 2015]
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.
⇒
x^{2} = 16
⇒ x = 3
(OR + RT)^{2} = 25 + PT^{2}
(3 + y)^{2} = 25 + PT^{2}
9 + y^{2} + 6y = 25 + PT^{2} .......(i)
In ΔPRT, TP^{2} = PR^{2} + RT^{2}
⇒ PT^{2} = (4)^{2} + (y)^{2} .......(ii)
Put value of PT^{2} in eq (i)
9 + y^{2} + 6y = 25 + 16 + y^{2}
6y = 25 + 16  9
6y = 32
y = 32/6 = 16/3 cm
Putting y = 16/3 cm in eq (ii), we get
Q40: If from an external point P of a circle with centre 0, two tangents PQ and PR are drawn such that QPR = 120°, prove that 2PQ = PO. [CBSE Delhi (F) 2016]
View AnswerAns: 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.)
Q41: In Fig. 8.42, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APS = 60°. Find the length of chord AB. [CBSE Delhi 2016]
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.
Q42: In Fig. 8.43 from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°. [CBSE (AI) 2016]
Ans: Let ∠TOP = θ
∴
Hence, ∠TOS = 120°
In ∠OTS, OT = OS (Radii of circle)
⇒
Q43: In Fig. 8.44, are two concentric circles of radii 6 cm and 4 cm with centre O. If AP is a tangent to the larger circle and BP to the smaller circle and length of AP is 8 cm, find the length of BP. [CBSE (F) 2016]
Ans: OA = 6 cm, OB = 4 cm, AP = 8 cm
OP^{2} = OA^{2} + AP^{2} = 36 + 64 = 100
⇒ OP = 10 cm
BP^{2} = OP^{2}  OB^{2} = 100  16 = 84
⇒
Q44: From an external point P, tangents PA and PR are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB. [CBSE Delhi 2016]
Ans: ∵ PA = PB ⇒ ∠BAP = ∠ABP = 50°
∴ ∠APB = 180°  50°  50° = 80°
and ∠AOB = 180°  80° = 100°
Q45: In Fig. 8.29, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA. [CBSE (AI) 2016]
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°
Q46: Two tangents PA and PB are drawn to the circle with centre O, such that ∠APB = 120°. Prove that OP = 2AP. (Foreign 2016)
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°
Q47: In Fig. 8.62, two equal circles, with centres O and O', touch each other at X.OO' produced meets the circle with centre O' at A. AC is tangent to the circle with centre O, at the point C. O'D is perpendicular to AC. Find the value of [CBSE (AI) 2016]
Ans: AC is tangent to circle with centre O.
Thus ∠ACO = 90°
In ΔAO'D and ΔAOC
∠ADO' = ∠ACO = 90º
∠A = ∠A (Common)
Q48: In Fig. 8.63, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle. [CBSE Delhi 2016]
Ans:
(Tangents from an external point to a circle are equal)
In right ΔAET.
TA^{2} = TE^{2} + EA^{2}
⇒ (12  x)^{2} = 64 + x^{2} ⇒ 144 + x^{2}  24x = 64 + x^{2}
⇒ x = 80/24 ⇒ x = 3.3 cm
Thus, AB = 6.6 cm
Q49: Prove that the parallelogram circumscribing a circle is a rhombus. [CBSE Delhi 2014; CBSE 2019 (30/5/1)]
View AnswerAns: 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.
Q50: Prove that the lengths of two tangents drawn from an external point to a circle are equal. [CBSE, Delhi 2014, (F) 2014, Delhi 2016, (AI) 2016, (F) 2016, CBSE Delhi 2017, (AI) 2017, (F) 2017, Delhi 2017 (C)]
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 RHScriterion of congruence, we get
Hence, lengths of two tangents from an external point are equal.
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