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.
Q16: 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
Q17: 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
Q18: 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.)
Q19: 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.
Q20: 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)
⇒
Q21: 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
⇒
Q22: 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°
Q23: 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°
Q24: 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°
Q25: 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)
Q26: 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
Q27: 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.
Q28: 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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