Short Answer Type Questions(Part- 2)- Circles Class 10 Notes | EduRev

Class 10 Mathematics by VP Classes

Class 10 : Short Answer Type Questions(Part- 2)- Circles Class 10 Notes | EduRev

The document Short Answer Type Questions(Part- 2)- Circles Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- 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(Part- 2)- Circles Class 10 Notes | EduRev

Short Answer Type Questions(Part- 2)- 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(Part- 2)- Circles Class 10 Notes | EduRev

Short Answer Type Questions(Part- 2)- Circles Class 10 Notes | EduRev

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

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

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