The document Very Short Answer Type Questions- Circles Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

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**VERY SHORT ANSWER TYPE QUESTIONS**

**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

OP^{2} = OT^{2} + PT^{2}

â‡’ 13^{2} = OT^{2} + 12^{2}

â‡’ OT^{2} = 13^{2} âˆ’ 12^{2} = (13 âˆ’ 12)

(13 + 12) = 1 Ã— 25 = 25

âˆ´ OT^{2} = 5^{2}

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

OP^{2} = OT^{2} + PT^{2}

= 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

OB^{2} = OA^{2} + AB^{2}

â‡’ 10^{2} = 6^{2} + AB^{2}

â‡’ AB^{2} = 10^{2} âˆ’ 6^{2} = (10 âˆ’ 6) (10 +6) = 4 Ã— 16 = 64 = 8^{2}

**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°

Now, in quadrilateral OAPB,

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:

OA^{2} + AP^{2} = OP^{2}

â‡’ r^{2} + (15)^{2} = (17)^{2}

r^{2} = 17^{2} âˆ’ 15^{2} = (17 âˆ’ 15) (17 + 15)

= 2 Ã— 32 = 64

Thus, radius = 8 cm.

**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°

In quadrilateral OABP,

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°

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