Short Answer Type Questions: Circles Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : Short Answer Type Questions: Circles Class 10 Notes | EduRev

The document Short Answer Type Questions: Circles Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

SHORT ANSWER TYPE QUESTIONS

Q1. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Sol.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Let NM be a chord of a circle with centre C.
Let the tangents at M and N meet at O.
∵ OM is a tangent at M

∵ ∠OMC = 90° ...(1)
Similarly ∠ONC = 90° ...(2)
Since, CM = CN [Radii of the same circle]
∵ In D CMN, ∠1= ∠2
From (1) and (2), we have
∠OMC –∠1 = ∠ONC –∠2
⇒ ∠OML = ∠ONL

Thus, tangents make equal angles with the chord.

Q2. Two concentric circles have a common centre O. The chord AB to the bigger circle touches the smaller circle at P. If OP = 3 cm and AB = 8 cm then find the radius of the bigger circle.

Sol. ∵ AB touches the smaller circle at P.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

∴ OP ⊥ AB ⇒ ∠OPA = 90°
Now, AB is a chord of the bigger circle.
Since, the perpendicular from the centre to a chord, bisects the chord,
∴ P is the mid-point of AB

⇒ Short Answer Type Questions: Circles Class 10 Notes | EduRev
In right ∆ APO, we have
AO= OP2 + AP2
⇒ AO2 = 32 + 42
⇒ AO2 = 9 + 16 = 25 = 52

⇒ Short Answer Type Questions: Circles Class 10 Notes | EduRev

Thus, the radius of the bigger circle is 5 cm.

Q3. In the given figure, O is the centre of the circle and PQ is a tangent to it. If its circumference is 12π cm, then find the length of the tangent.

Sol. ∵ Circumference of the circle = 12π cm

Short Answer Type Questions: Circles Class 10 Notes | EduRev

∴ 2π r =12π
[∵ r is the radius of the circle]

⇒  Short Answer Type Questions: Circles Class 10 Notes | EduRev
⇒ Radius of the circle = 6 cm = OQ

Since a tangent to circle is perpendicular to the radius through the point of contact,
∴ ∠OQP = 90°
Now, in rt Δ OQP, we have:
OQ2 + QP2 = OP2
⇒ 62 + QP2 =102
⇒ QP2 = 102 − 62 = (10 − 6) (10 + 6) = 4 × 16 = 64 = 82

⇒  Short Answer Type Questions: Circles Class 10 Notes | EduRev

Thus, the length of the tangent is 8 cm.

Q4. Given two concentric circles of radii 10 cm and 6 cm. Find the length of the chord of the larger circle which touches the other circle.

Sol. The chord AB touches the inner circle at P.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

∴ AB is tangent to the inner circle.
⇒ OP ⊥ AB

[∵ O is the centre and OP is radius through the point of contact P]
∴ ∠OPB = 90°.
Now, in right ∆ OPB, we have:
OP2 + PB2 = OB2
⇒ 62 + PB2 = 102
⇒ PB2 = 102 − 62
= (10 − 6) × (10 + 6)
⇒ PB2 = 4 × 16
⇒ PB2 = 64 = 82

⇒ Short Answer Type Questions: Circles Class 10 Notes | EduRev

∵ The radius perpendicular to a chord bisects the chord.
∴ P is the mid-point of AB
∴ AB = 2 × PB = 2 × 8 = 16 cm.

Q5. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. ∵ Tangent to a circle is perpendicular to the radius through the point of contact.
In quadrilateral. OPTQ,
∠OPT + ∠OQT + ∠POQ + ∠PTQ = 360°
or 90° + 90° + ∠POQ + ∠PTQ = 360°
⇒ ∠POQ + ∠PTQ = 360° − 90° − 90° = 180°   ...(1)
In Δ OPQ, ∠1 + ∠2 + ∠POQ = 180°   ...(2)

Since OP =OQ   [Radii of the same circle]
⇒ ∠1 = ∠2    [Angles opposite to equal sides]
∴  ∠OPT = 90° = ∠OQT
∴ From (2), we have
∠1 + ∠1 + ∠POQ = 180°
⇒ 2 ∠1 + ∠POQ = 180° ...(3)
From (1) and (3), we have
2 ∠1 + ∠POQ = ∠POQ + ∠PTQ
⇒ 2 ∠1= ∠PTQ
⇒ 2 ∠OPQ = ∠ PTQ.

Q6. In the figure, the incircle of ∆ ABC touches the sides BC, CA and AB at D, E and F respectively. If AB = AC, prove that BD = CD.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. Since the lengths of tangents drawn from an external point to a circle are equal,
∴ We have
AF = AE
BF = BD
CD = CE
Adding them, we get
(AF + BF) + CD = (AE + CE) + BD
⇒ AB + CD = AC + BD
But AB = AC  (Given)
∴ CD = BD.

Q7. A circle is touching the side BC of a D ABC at P and touching AB and AC produced at Q and R. Prove that:  Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. Since, the two tangents drawn to a circle from an external point are equal.
∵  AQ = AR ...(1)
Similarly, BQ = BP ...(2)
and CR = CP ...(3)

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Now, Perimeter of Δ ABC
= AB + BC + AC
= AB + (BP + PC) + AC
= AB + (BQ + CR) + AC
[From (2) and (3)]
= (AB + BQ) + (CR + AC)
= AQ + AR
= AQ + AQ [From (1)] = 2AQ
⇒ Short Answer Type Questions: Circles Class 10 Notes | EduRev

Q8. In the given figure, the radii of two concentric circles are 13 cm and 8 cm. AB is diameter of the bigger circle. BD is the tangent to the smaller circle touching it at D. Find the length AD. 

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. Join AE and OD
∵ ∠ODB = 90°
[BE is a tangent at D and OD is a radius] and ∠AEB = 90° [AB is diameter so ∠AEB is an angle in semicircle so ∠AEB = 90°]
∴ OD ║ ΑΕ and ∆BEA are similar by AA-Similarity

So Short Answer Type Questions: Circles Class 10 Notes | EduRev
Short Answer Type Questions: Circles Class 10 Notes | EduRev

Again in rt  Δ AED,

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Q9. In two concentric circles, a chord of the larger circle touches the smaller circle. If the length of this chord is 8 cm and the diameter of the smaller circle is 6 cm, then find the diameter of the larger circle.

Sol. Let the common centre be O. Let AB be the chord of the larger circle.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

∴ AB = 8 cm

And CD is the diameter of the smaller circle i.e., CD = 6 cm

⇒  Short Answer Type Questions: Circles Class 10 Notes | EduRev
Join OA. D is the point of contact.
∴ OD ⊥ AB
⇒ D is the mid point of AB
⇒ AD = 4 cm Now, in right ΔADO, we have:
AO2 = AD2 + OD2
= 42 + 32 = 16 + 9 = 25 = 52
⇒ AO = 5 cm
⇒ 2AO = 2(5 cm) = 10 cm

∴ The diameter of the bigger circle is 10 cm.

Q10. In the following figure, PA and PB are two tangents drawn to a circle with centre O, from an external point P such that PA = 5 cm and ∠APB = 60°. Find the length of chord AB.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. Since the tangents to a circle from an external point are equal,
∴ PA = PB = 5 cm
In ΔPAB, we have
∠PAB = ∠PBA               [∵ PA = PB]
∴∠PAB + ∠PBA + ∠APB = 180°
⇒∠PAB + ∠PAB + 60° = 180°
⇒ 2 ∠PAB + 60° = 180°
⇒ 2 ∠PAB = 180°− 60°
= 120°
⇒ ∠PAB = 60°
⇒ Each angle of ∆PAB is 60°.
⇒ ΔPAB is an equilateral triangle. ∴
PA = PB
= AB = 5 cm
Thus, AB = 5 cm

Q11. In the following figure, AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

The tangents at A and B intersect at P. Find the length PA.

Sol.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Join OB.
Let PA = x cm and PR = y cm Since, OP is perpendicular bisector of AB

∴ AR = BR = 9.6/2 = 4.8 cm

Now, in rt ∆OAR, we have:
OA2 = OR2 + AR2
[By Pythagoras theorem]
⇒ OR2 = OA2 − AR2
= 62 − (4.8)2 = (6 − 4.8) × (6 + 4.8) = 1.2 × 10.8
⇒ = 12.96
OR = 3.6 cm.
Again, in right ΔOAP,
OP2 = AP2 + OA2
OP2 = (AR2 + PR2) + OA2
[∵ AP2 = AR2 + PR2]
⇒ (y + 3.6)2 = (4.8)2 + y2 + 62
⇒y2 + 12.96 + 7.2 y = 23.04 + y2 + 36
⇒ 7.2 y = 46.08

⇒  Short Answer Type Questions: Circles Class 10 Notes | EduRev
⇒ PR = 6.4 cm
Now, AP2 = AP2 + PR2
= (4.8)2 + (6.4)2 = 23.04 + 40.96 = 64

⇒  Short Answer Type Questions: Circles Class 10 Notes | EduRev


Q12. Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove  that ∠APB = 2∠OAB

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. We have PA and PB, the tangents to the circle and O is the centre of the circle.
∴ PA = PB
⇒∠2 = ∠4  ...(1)

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Since the tangent is perpendicular to the radius through the point of contact,
∴∠OAP = 90°
⇒ ∠1 + ∠2 = 90° ...(2)
⇒ ∠2 = 90° − ∠1
Now, in ∆ABP, we have:
∴∠2 + ∠3 + ∠4 = 180°
⇒∠2 + ∠3 + ∠2 = 180° [From (1)]
⇒∠2 + ∠3 = 180°
⇒ 2 (90° − ∠1) + ∠3 = 180° [From (2)]
⇒ 180° − 2 ∠1 + ∠3 = 180°
⇒ 2 ∠1 = ∠3
⇒ ∠3 = 2∠1
⇒∠APB = 2∠OAB

Q13. ABC is an isosceles triangle, in which AB = AC, circumscribed about a circle. Show that BC is bisected at the point of contact.

Sol. We know that the tangents to a circle from an external point are equal.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

∴ AD = AF
Similarly,
BD = BE
and CE = CF
Since AB = AC [Given]
⇒ AB − AD = AC − AD
⇒ AB − AD = AC − AF [∵ AD = AF]
⇒ BD = CF ...(1)
But BF = BD and CF = CE
∴ From (1), we have:
BE = CE

Q14. If a, b, c are the sides of a right triangle where c is hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by  Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. Here, a, b and c are the sides of rt D ABC such that BC = a, CA = b and AB = c Let the circle touches the sides BC, CA, AB at D, E and F respectively.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

= AE = AF and BD = BF
Also, CE = CD = r
∴ AF = b – r BF = a – r
Now, AB = c  ⇒  (AF + BF)
=(b – r) + (a – r)
⇒ c = b + a – 2r
⇒ 2r = a + b – c

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Q15. In a right Δ ABC, right angled at B, BC = 5 cm and AB = 12 cm. The circle is touching the sides of Δ ABC. Find the radius of the circle.

Sol. Let the circle with centre O and radius ‘r’ touches AB, BC and AC at P, Q, R, respectively.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Now,
AR = AP
∵ AP = AB – BP = (12 – r) cm
∴ AR = (12– r)cm
Similarly, CR = (5 – r)cm
Now, using Pythagoras theorem in rt Δ ABC, we have
AC2 = AB2 + BC2
⇒ AC2 = 122 + 52
⇒ AC = 13 cm
But AC = AR + CR = (12 – r) + (5 – r)
⇒ (12 – r) + (5 – r) = 13 cm
⇒ 17 – 2r = 13 cm
⇒ 2 r = 17 – 13 = 4 cm
⇒ r = 4/2 = 2 cm

Thus, the radius of the circle is 2 cm.

Q16. Prove that the parallelogram circumscribing a circle is a rhombus.

Sol. Since ABCD is a ║ gm
∴ AB = CD
and AD = BC
∵ Tangents from an external point to a circle are equal,

Short Answer Type Questions: Circles Class 10 Notes | EduRev

⇒(AP + PB) + (RC + DR)
  = (AS + DS) + (BQ + QC)
⇒ AB + CD = AD + BC
⇒ 2 AB =2 AD ⇒ AB = AD
⇒ AB = AD = CD = BC
i.e., ABCD is a rhombus.

Q17. In the following figure, OP is equal to diameter of the circle. Prove that ABP is an equilateral triangle.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. Since the tangent is perpendicular to the radius through the point of contact,
∴ ∠OAP = 90°

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Let us join AB and AC.
In right ∆OAP, OP is the hypotenuse and C is the mid point of OP.
[∵ OP is a diameter of the circle (given)]
∴ CA = CP = CO = Radius of the circle.
∴ ΔOAC is an equilateral triangle.
Since all angles in an equilateral triangle are 60°,
∴∠1 = 60°
Now, in ∆OAP, we have
∠1 + ∠OAP + ∠2 = 180°
⇒ 60° + 90° + ∠2 = 180°
⇒∠2 = 180° − 90° − 60° = 30°
Since PA and PB make equal angles with OP,
∴∠2= ∠3 ⇒∠3 = 30°
∴∠APB = ∠2 + ∠3
= 30° + 30° = 60°
Again, PA = PB.
⇒ In  ΔABP,∠4= ∠5
[Angles opposite to equal sides are equal]
Now, in ΔABP, ∠4 + ∠5 + ∠APB = 180°
⇒∠4 + ∠4 + ∠APB= 180°
⇒ 2∠4 + ∠60° = 180°
⇒ 2∠4 = 180° − 60° = 120°

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Since, ∠4 = 60° ∠5 = 60°
∵ ΔABP is an equilateral Δ.
∠APB = 60°

Q18. Prove that the angle between the two tangents to a circle drawn from an external point is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Or

Two tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that AOBP is a cyclic quadrilateral.

Sol. 

Short Answer Type Questions: Circles Class 10 Notes | EduRev

We have tangents PA and PB to the circle from the external point P. Since a tangent to a circle is perpendicular to the radius through the point of contact,
∴ ∠2 = 90° and ∠4 = 90°
Now, in quadrilateral OAPB,
∠1 + ∠2 + ∠3 + ∠4 = 360°
⇒∠1 + 90° + ∠3 + 90° = 360°
⇒∠1 + ∠3 = 360° − 90° − 90°
= 180°
i.e., ∠1 and ∠3 are supplementary angles.
⇒∠ AOB and ∠APB are supplementary
⇒ AOBP is a cyclic quadrilateral.

Q19. 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  Short Answer Type Questions: Circles Class 10 Notes | EduRev

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. AC is tangent to circle with centre O at C (given)
∠ACO = 90°  ⇒ ∆ ACO is a rt Δ
∠ADO′ = 90°   [ ä O′D ⊥ AC] ⇒ Δ
ADO′ is a rt Δ

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Short Answer Type Questions: Circles Class 10 Notes | EduRev

But, AO′ = r,     O′X = r  and  OX = r ⇒ AO = 3r

Short Answer Type Questions: Circles Class 10 Notes | EduRev
From (1) and (2), we get  Short Answer Type Questions: Circles Class 10 Notes | EduRev

Q20. Out of two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. Let the given chord AC of the larger circle touch the smaller circle at L. ∵ AC is a tangent at L to the smaller circle with centre O
∴ OL ⊥ AC
Also AC is a chord of the bigger circle

∴  Short Answer Type Questions: Circles Class 10 Notes | EduRev

Now, in rt. ΔOAL,
OL2 = OA2 – AL2

or OL2 = 52 – 42
= (5 + 4) (5 – 4)
= 9 × 1 = 9

⇒  Short Answer Type Questions: Circles Class 10 Notes | EduRev

Thus, the radius of the inner circle is 3 cm.

Q21. In the figure, O is the centre of a circle of radius 5cm. T is a point such that OT = 13cm 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. 

Short Answer Type Questions: Circles Class 10 Notes | EduRev

Sol. O is centre of the circle and PT is a tangent to circle
∴ ∠ OPT = 90°  ⇒ ∆ OPT is a rt ∆ using Pythagoras theorem
OT2 = OP2 + PT2   or 132 = 52 + PT2
⇒ PT2 =132 – 52  
⇒ PT =  Short Answer Type Questions: Circles Class 10 Notes | EduRev

Let AP = AE = x
[Tangent to a circle from an external point are equal]
⇒ AT = PT – AP = (12 – x) cm
[∵ AB is a tangent to the circle at E and OE is a radius]
∴ ∠ OEA = 90° ⇒ ∠AET  = 90°
Δ AET is a rt ∆
⇒ AT2 = AE2 + ET2
or (12 − x)2 = x2 + (13 − 5)2
⇒  144− 24x + x2 = x2 + 64 or  24x = 80

Short Answer Type Questions: Circles Class 10 Notes | EduRev

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