Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Long Questions: Circles

Class 10 Maths Chapter 10 Question Answers - Circles

Q1: Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.
Ans:

Class 10 Maths Chapter 10 Question Answers - CirclesGiven: CD and EF are two C parallel tangents at points A and B of a circle with centre O.
To prove: AB passes through centre O or AOB is diameter of the circle.
Const.: Join OA and OB. Draw OM || CD.
Proof: ∠1 = 90° … (i)
…[∵ Tangent is I to the radius through the point of contact
OM || CD
∴ ∠1 + ∠2 = 180° …(Co-interior angles
90° + ∠2 = 180° …[From (i)
∠2 = 180° – 90o = 90°
Similarly, ∠3 = 90°
∠2 + ∠3 = 90° + 90° = 180°
∴ AOB is a straight line.
Hence AOB is a diameter of the circle with centre O.
∴ AB passes through centre 0. 

Q2: In the figure, the sides AB, BC and CA of triangle ABC touch a circle with centre o and radius r at P, Q and R. respectively.
Prove that:
(i) AB + CQ = AC + BQ
(ii) Area (AABC) = 1/2 (Perimeter of ∆ABC ) × r

Class 10 Maths Chapter 10 Question Answers - Circles

Ans:
Class 10 Maths Chapter 10 Question Answers - CirclesPart I:
Proof: AP = AR …(i)
BP = BQ … (ii)
CQ = CR … (iii)
Adding (i), (ii) & (iii)
AP + BP + CQ
= AR + BQ + CR
AB + CQ = AC + BQ
Part II: Join OP, OR, OQ, OA, OB and OC
Proof: OQ ⊥ BC; OR ⊥ AC; OP ⊥ AB
ar(∆ABC) = ar(∆AOB) + ar(∆BOC) + ar (∆AOC)
Area of (∆ABC)
Class 10 Maths Chapter 10 Question Answers - Circles

Q3: In the figure, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line I at D and m at E. Prove that ∠DOE = 90°.
Class 10 Maths Chapter 10 Question Answers - CirclesAns: Proof: Let I be XY and m be XY’
∠XDE + ∠X’ED = 180° … [Consecutive interior angles]
Class 10 Maths Chapter 10 Question Answers - CirclesClass 10 Maths Chapter 10 Question Answers - Circles
= ∠1 + ∠2 = 90° …[OD is equally inclined to the tangents
In ∆DOE, ∠1 + ∠2 + 23 = 180° …[Angle-sum-property of a ∆
90° + 23 = 180°
⇒ ∠3 = 180° – 90o = 90°
∴ ∠DOE = 90° …(proved) 

Q4: In the figure, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line I at D and m at E. Prove that ∠DOE = 90°.
Class 10 Maths Chapter 10 Question Answers - CirclesAns: Proof: Let I be XY and m be XY’
∠XDE + ∠X’ED = 180° … [Consecutive interior angles]
Class 10 Maths Chapter 10 Question Answers - CirclesClass 10 Maths Chapter 10 Question Answers - Circles
= ∠1 + ∠2 = 90° …[OD is equally inclined to the tangents
In ∆DOE, ∠1 + ∠2 + 23 = 180° …[Angle-sum-property of a ∆
90° + 23 = 180°
⇒ ∠3 = 180° – 90o = 90°
∴ ∠DOE = 90° …(proved) 

Q5: Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Ans:

Class 10 Maths Chapter 10 Question Answers - CirclesB is the mid point of arc (ABC)
OA = OC …[Radius
OF = OF …[Common
∴ ∠1 = ∠2 …[Equal angles opposite equal sides
∴ ∆OAF = ∆OCF (SAS)
∴ ∠AFO = ∠CFO = 90° …[c.p.c.t
⇒ ∠AFO = ∠DBO = 90° …[Tangent is ⊥to the radius through the point of contact
But these are corresponding angles,
∴ AC || DE 


Q6: Prove that the lengths of tangents drawn from an external point to a circle are equal.
Ans: Given: PT and PS are tangents from an external point P to the circle with centre O.
Class 10 Maths Chapter 10 Question Answers - CirclesTo prove: PT = PS
Const.: Join O to P,
T & S
Proof: In ∆OTP and
∆OSP,
OT = OS …[radii of same circle
OP = OP …[circle
∠OTP – ∠OSP …[Each 90°]
∴ AOTP = AOSP …[R.H.S]
PT = PS …[c.p.c.t]

Q7: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Ans: 

Class 10 Maths Chapter 10 Question Answers - CirclesGiven: XY is a tangent at point P to the circle with centre O.
To prove: OP ⊥ XY
Const.: Take a point Q on XY other than P and join to OQuestion
Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle.
∴ OQ > OP
This happens with every point on the line XY except the point P.
OP is the shortest of all the distances of the point O to the points of XY
∴ OP ⊥ XY … [Shortest side is ⊥]

Q8: 
Class 10 Maths Chapter 10 Question Answers - Circles In the above figure, PQ is a chord of length 16 cm, of a circle of radius 10 cm. The tangents at P and Q intersect at a point T. Find the length of TP.

Ans:
TP = TQuestion .. [Tangents drawn from an external point]
∆TPQ is an isosceles ∆ and TO is the bisector of ∠PTQ ,
OT ⊥ PQ …[Tangent is ⊥ to the radius through the point of contact
∴ OT bisects PQ
∴ PR = RQ = 16 = 8 cm …[Given]
In rt. ∆PRO,
PR2 + RO2 = PO2 … [Pythagoras’ theorem]
82 + RO2 = (10)2
RO2 = 100 – 64 = 36
∴ RO = 6 cm
Let TP = x cm and TR = y cm
Then OT = (y + 6) cm
In rt. ∆PRT, x2 = y2 + 82 …(i) …[Pythagoras’ theorem]
In rt. ∆OPT,
OT2 = TP2 + PO2 …(Pythagoras’ theorem]
(y + 6)2 = x2 + 102
y2 + 12y + 36 = y2 +64 + 100 …[From (i)]
Class 10 Maths Chapter 10 Question Answers - Circles

Q9: In the figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQuestion Find ∠RQS.
Class 10 Maths Chapter 10 Question Answers - CirclesAns:

PR = PO …[∵ Tangents drawn from an external point are equal
⇒ ∠PRQ = ∠PQR …[∵ Angles opposite equal sides are equal
In ∆PQR, 
Class 10 Maths Chapter 10 Question Answers - Circles⇒ ∠PRQ + ∠RPQ + ∠POR = 180°…[∆ Rule]
⇒ 30° + 2∠PQR = 180° 
Class 10 Maths Chapter 10 Question Answers - Circles
= 75°
⇒ SR || QP and QR is a transversal
∵ ∠SRQ = ∠PQR … [Alternate interior angle
∴ ∠SRO = 75° …..[Tangent is I to the radius through the point of contact
⇒ ∠ORP = 90°
∴ ∠ORP = ∠ORQ + ∠QRP
90° = ∠ORQ + 75°
∠ORQ = 90° – 75o = 150
Similarly, ∠RQO = 15°
In ∆QOR,
∠QOR + ∠QRO + ∠OQR = 180° …[∆ Rule
∴∠QOR + 15° + 15° = 180°
∠QOR = 180° – 30° = 150° 
Class 10 Maths Chapter 10 Question Answers - Circles
In ARSQ, ∠RSQ + ∠QRS + ∠RQS = 180° … [∆ Rule]
∴ 75° + 75° + ∠RQS = 180°
∠RQS = 180° – 150° = 30° 

Q10: In the figure, 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.
Class 10 Maths Chapter 10 Question Answers - CirclesAns:
∠OPT = 90° …[Tangent is ⊥ to the radius through the point of contact
We have, OP = 5 cm, OT = 13 cm
In rt. ∆OPT,
OP2 + PT2 = OT? …[Pythagoras’ theorem
⇒ (5)2 + PT2 = (13)2
⇒ PT2 = 169 – 25 = 144 cm
⇒ PT = √144144
= 12 cm
OP = OQ = OE = 5 cm … [Radius of the circle
ET = OT – OE
= 13 – 5 = 8 cm
Let, PA = x cm, then AT = (12 – x) cm
PA = AE = x cm …[Tangent drawn from an external point
In rt. ∆AET,
AE2 + ET2 = AT2 …(Pythagoras’ theorem
⇒ x2 + (8)2 = (12 – x)2
⇒ x2 + 64 = 144 + x2 – 24x
⇒ 24x = 144 – 64  

Class 10 Maths Chapter 10 Question Answers - Circles

The document Class 10 Maths Chapter 10 Question Answers - Circles is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
124 videos|457 docs|77 tests

Up next

FAQs on Class 10 Maths Chapter 10 Question Answers - Circles

1. What is the formula for the area of a circle?
Ans. The formula for the area of a circle is given by \( A = \pi r^2 \), where \( A \) is the area and \( r \) is the radius of the circle.
2. How do you find the circumference of a circle?
Ans. The circumference of a circle can be found using the formula \( C = 2\pi r \) or \( C = \pi d \), where \( C \) is the circumference, \( r \) is the radius, and \( d \) is the diameter of the circle.
3. What is the relationship between the radius and diameter of a circle?
Ans. The diameter of a circle is twice the length of its radius. This can be expressed as \( d = 2r \), where \( d \) is the diameter and \( r \) is the radius.
4. How do you calculate the length of an arc in a circle?
Ans. The length of an arc can be calculated using the formula \( L = \frac{\theta}{360} \times 2\pi r \), where \( L \) is the length of the arc, \( \theta \) is the angle in degrees subtended by the arc at the center of the circle, and \( r \) is the radius.
5. What is the definition of a tangent to a circle?
Ans. A tangent to a circle is a straight line that touches the circle at exactly one point. This point is called the point of tangency, and the tangent is perpendicular to the radius drawn to this point.
124 videos|457 docs|77 tests
Download as PDF

Up next

Explore Courses for Class 10 exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Viva Questions

,

Free

,

MCQs

,

mock tests for examination

,

study material

,

Extra Questions

,

Summary

,

Class 10 Maths Chapter 10 Question Answers - Circles

,

practice quizzes

,

past year papers

,

Class 10 Maths Chapter 10 Question Answers - Circles

,

video lectures

,

pdf

,

shortcuts and tricks

,

Previous Year Questions with Solutions

,

Sample Paper

,

ppt

,

Objective type Questions

,

Exam

,

Semester Notes

,

Important questions

,

Class 10 Maths Chapter 10 Question Answers - Circles

;