Courses

# Very Short Answer Type Questions- Circles Class 10 Notes | EduRev

## Class 10 : Very Short Answer Type Questions- Circles Class 10 Notes | EduRev

The document Very 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

Q1. In the adjoining figure, PA and PB are tangents from P to a circle with centre C. If âˆ APB = 40Â° then find âˆ ACB.

Sol. Since a tangent to a circle is perpendicular to the radius through the point of contact,

âˆ´ âˆ 1 = 90Â° and âˆ 2 = 90Â°
Now, in quadrilateral APBC, we have:
âˆ 1 + âˆ ACB + âˆ 2 + âˆ P = 360Â°
â‡’90Â° + âˆ ACB + 90Â° + 40Â°= 360Â°
â‡’âˆ ACB + 220Â° = 360Â°
â‡’âˆ ACB = 360Â° âˆ’ 220Â°
= 140Â°.

Q2. In the given figure, PA and PB are tangents from P to a circle with centre O. If âˆ AOB = 130Â°, then find âˆ APB.

Sol. Since a tangent to a circle is perpendicular to the radius through the point of contact,

âˆ´âˆ 1= âˆ 2 = 90Â°
Now, in quadrilateral AOBP, we have:
âˆ 1 + âˆ AOB + âˆ 2 + âˆ APB = 360Â°
â‡’ 90Â° + 130Â° + 90Â° + âˆ APB = 360Â°
â‡’ 310Â° + âˆ APB = 360Â°
â‡’âˆ APB = 360 âˆ’ 310
= 50Â°
Thus, âˆ APB = 50Â°.

Q3. In the given figure, PT is a tangent to a circle whose centre is O. If PT = 12 cm and PO = 13 cm then find the radius of the circle.

Sol. Since a tangent to a circle is perpendicular to the radius through the point of contact,
âˆ´âˆ OTP =90Â°
In rt Î” OTP, using Pythagoras theorem, we get

OP2 = OT2 + PT2
â‡’ 132 = OT2 + 122
â‡’ OT2 = 132 âˆ’ 122 = (13 âˆ’ 12)
(13 + 12) = 1 Ã— 25 = 25
âˆ´ OT2 = 52
â‡’ OT = 5
Thus, radius (r) = 5 cm.

Q4. In the given figure, PT is a tangent to the circle and O is its centre. Find OP.

Sol. Since, a tangent to a circle is perpendicular to the radius through the point of contact.
âˆ´âˆ OTP = 90Â°
In right Î” OTP, using Pythagoras theorem, we get
OP2 = OT2 + PT2
= 82 + 152 = 64 + 225 = 289 = 172

Q5. If O is the centre of the circle, then find the length of the tangent AB in the given figure.

Sol. âˆµ A tangent to a circle is perpendicular to the radius through the point of contact.
âˆ´ âˆ OAB = 90Â°
Now, in right Î” OAB, we have
OB2 = OA2 + AB2
â‡’ 102 = 62 + AB2
â‡’ AB2 = 102 âˆ’ 62 = (10 âˆ’ 6) (10 +6) = 4 Ã— 16 = 64 = 82

Q6. In the figure, PA is a tangent from an external point P to a circle with centre O. If âˆ POB = 115Â° then find âˆ APO.

Sol. Here, PA is a tangent and OA is radius.
Also, a radius through the point of contact is perpendicular to the tangent.
âˆ´ OA = PA
â‡’âˆ PAO = 90Â°
In âˆ†OAP, âˆ POB is an external angle,
âˆ´âˆ APO + âˆ PAO = âˆ POB
â‡’âˆ  APO + 90Â° = 115Â°
â‡’ âˆ APO = 115Â° âˆ’ 90Â° = 25Â°

Q7. In the following figure, PA and PB are tangents drawn from a point P to the circle with centre O. If âˆ APB = 60Â°, then what is âˆ AOB?

Sol. The radius of the circle through the point of contact is perpendicular to the tangent.

âˆ´ OA âŠ¥ AP and OB âŠ¥ BP
â‡’ âˆ PAO = âˆ PBO = 90Â°
âˆ OAP + âˆ APB + âˆ PBO + âˆ AOB = 360Â°
90Â° + 60Â° + 90Â° + âˆ AOB = 360Â°
â‡’ âˆ AOB + 240Â° = 360Â°
â‡’âˆ AOB = 360Â° â€“ 240Â°
= 120Â°

Q8. In the figure, CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If QC = 11 cm, BC = 7 cm then find, the length of BR.

Sol. âˆµ Tangents drawn from an external point are equal,
âˆ´ BQ = BR ...(1)
And CQ = CP
Since, BC + BQ = QC
â‡’ 7 + BR =11 [âˆµ BQ = BR]
BR = 11 âˆ’ 7 = 4 cm.

Q9. In the figure, Î”ABC is circumscribing a circle. Find the length of BC.

Sol. Since tangents drawn from an external point to the circle are equal,
âˆ´ AR = AQ = 4 cm ...(1)
BR = BP = 3 cm ...(2)
PC = QC ...(3)
âˆ´ QC = AC âˆ’ AQ
= 11 âˆ’ 4 = 7 cm [From (1)]
BC = BP + PC [From (3)]
= 3 + QC
= (3 + 7) cm = 10 cm

Q10. In the figure, if âˆ  ATO = 40Â°, find âˆ  AOB.

Sol. Since the tangent is perpendicular to the radius through the point of contact,
âˆ´ âˆ 1= âˆ 4 = 90Â°
Also, OA = OB    [Radii of the same circle]
OT = OT      [Common]
âˆ´âˆ†OAT â‰…âˆ†OBT      [RHS congruency]
â‡’âˆ 3= âˆ 2

Now, in Î”OAT,
âˆ 3 + âˆ 4 + âˆ 5 = 180Â°
â‡’ âˆ 3 + 90Â° + 40Â° = 180Â°
â‡’ âˆ 3 = 180Â° âˆ’ 90Â° âˆ’ 40Â° = 50Â°
â‡’ âˆ AOB = 50Â° + 50Â° = 100Â°.

Q11. From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm, then what is the radius of the circle?

Sol. Since radius is perpendicular to the tangent through the point of contact,
âˆ´ OA âŠ¥ AP
â‡’ âˆ OAP = 90Â°
In rt âˆ†OAP, we have:
OA2 + AP2 = OP2
â‡’ r2 + (15)2 = (17)2
r2 = 172 âˆ’ 152 = (17 âˆ’ 15) (17 + 15)
= 2 Ã— 32 = 64

Q12. The two tangents from an external point P to a circle with centre O are PA and PB. If âˆ APB = 70Â°, then what is the value of âˆ AOB?

Sol. Since tangent is perpendicular to the radius through the point of contact.

âˆ´âˆ 1= âˆ 2 = 90Â°
âˆ AOB + âˆ 1 + âˆ 2 + âˆ APB = 360Â°
âˆ AOB + 90Â° + 90Â° + 70Â° = 360Â°

â‡’âˆ AOB + 250Â° = 360Â°
â‡’âˆ AOB = 360Â°âˆ’250Â°
= 110Â°

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;