Class |
10 |

Chapter |
Circles |

Subject |
Math |

Category |
Important Question Answer |

**Class 10 Math Chapter 10 Important Question Answer**

**Q1. The lengths of tangents drawn from an external point to a circle are ___________.**

**Ans**. equal

**Q2. A circle can have _______ parallel tangents at the most. Most Important**

**Ans**. two

**Q3. All circles are _________. (Similar/Congruent) Most Important**

**Ans**. similar.

**Q4. A tangent to a circle intersects it in __________ points. Most Important**

**Ans**. one

**Q5. A line intersecting a circle in two points is called ___________. Most Important**

**Ans**. Secant

**Q6. A tangent to a circle intersects it in _________ point(s). **

**Ans.** one

**Q7. The common point of a tangent to a circle and the circle is called **** ****__________. Most Important**

**Ans**. point of contact

**Q8. From a point Q, the length of tangent to a circle is 24 cm and distance of Q from centre is 25 cm. Find the radius of circle.**

**Ans**.

Given, length of tangent PQ = 24 cm

OQ = 25 cm

Radius of Circle OP = ?

As we know, Tangent at any point on the circle is perpendicular to the radius through point of contact

Therefore, OT ⊥ PQ

In ∆ OPQ, Using Pythagoras theorem

OQ^{2} = OP^{2} + PQ^{2}

25^{2} = OP^{2} + 24^{2}

625 – 576 = 49 = OP^{2}

OP = 7 cm

Therefore, length of radius of given circle is 7cm.

**Q9. Find the length of tangent drawn from a point whose distance from the centre of a circle is 25 cm. Given the radius of circle is 7 cm.**

**Ans**.

Given, Radius of circle OT = 7 cm

PO = 25 cm

PT = ?

As we know, Tangent at any point on the circle is perpendicular to the radius through point of contact

Therefore, PT ⊥ OT

In ∆ PTO, Using Pythagoras theorem

PO^{2} = TO^{2} + PT^{2}

25^{2} = 7^{2} + PT^{2}

625 – 49 = 576 = PT^{2}

PT = 24 cm

Therefore, required length of tangent is 24 cm.

**Q10. PQ is a chord of length 8 cm of a circle of radius 5 cm tangents at P and Q intersect at a point T. Find the length TP, if O is the centre of a circle.**

**Ans**.

Join OT. Let it intersect PQ at the point R. Then ∆ TPQ is isosceles and TO is the angle bisector of ∠PTQ. So, OT⊥PQ and therefore, OT bisects PQ which gives PR = RQ = 4 cm.

Also, OR = = 3 cm

Now, ∠TPR + ∠RPO = 90° = ∠TPR + ∠PTR

So, ∠RPO = ∠PTR

Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.

This give, , i.e., or TP = cm.

**Note : TP can also be found by using pythagoras theorem. **

Let TP = x and TR = y. Then

Using Pythagoras theorem,

x^{2} = y^{2} + 16 (Taking right ∆ PRT) —— (i)

x^{2} + 5^{2} = (y + 3)^{2 }(Taking right ∆ OPT) —– (ii)

Subtracting eq (i) from (ii), we get

25 = 6y – 7

y =

putting value of y in eq (i), we get

x^{2} =

x = cm

**Q11. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.**

**Ans**.

Let us draw a circle with centre O and Diameter AB.

Let two tangents is drawn, one is PQ at point A and other is RS drawn at point B.

As we know,

OA ⊥ PQ (Tangent at any point on the circle is perpendicular to the radius through point of contact)

∠OAP = 90° —– (i)

Similarly, OB ⊥ RS (Tangent at any point on the circle is perpendicular to the radius through point of contact)

∠OBS = 90° —– (ii)

From eq (i) and (ii) we get,

∠OAP = ∠OBS = 90°

Since, AB is a transversal for lines PQ and RS and ∠BAP = ∠ABS

i.e. If alternate angles are equal then the line are parallel.

Therefore, PQ || RS.

Hence proved.

**Q12. Prove that the length of tangents drawn from an external point to a circle are equal. Most Most Important**

**Ans**.

We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P.

We are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then ∠OQP and ∠ORP are right angles, because these are angle between the radii and tangents, and they are right angles as the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Now in right triangles OQP and ORP,

OQ = OR (Radii of the same circle)

OP = OP (Common)

Therefore, OQP ≅ ORP (RHS)

This gives PQ = PR (CPCT)

**Q13. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that **∠**PTQ = 2**∠** OPQ.**

**Ans**.

We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the point of contact.

We need to prove that

∠PTQ = 2∠OPQ

Let ∠PTQ= θ

Now, TP = TQ as the lengths of tangents drawn from an external path to a circle are equal.

So, TPQ is an isosceles triangle.

Therefore, ∠TPQ = ∠TQP =

Also, ∠OPT = 90^{o} as the tangent at any point of a circle is perpendicular to the radius through the point of contact.

So, ∠OPQ = ∠OPT – ∠TPQ = 90^{o} –

=

This gives ∠PTQ = 2 ∠OPQ

**Q****14****. Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.**

**Ans**.

We are given two concentric circles, C_{1} and C_{2} with centre O and a chord AB of the larger circle C_{1} which touches the smaller circle C_{2} at the point P.

We need to prove that AP = BP.

Let us join OP.

Then, AB is a tangent to C_{2} at P and OP is its radius. Therefore, OP ⊥ AB

Now AB is a chord of the circle C_{1} and OP ⊥ AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the center bisects the chord,

i.e. AP = BP

**Q****15****. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.**

**Ans**.

Given radius of bigger circle OP = 5 cm

Radius of smaller circle OA = 3 cm

Find PQ = ?

OA ⊥ PQ (because radius of circle is always perpendicular to its tangent)

Also, PA = AQ (because ∆ OAP ≅ ∆ OAQ)

In ∆ OAP, Using pythagoras theorem,

PO^{2 }= PA^{2} + AO^{2}

5^{2} = PA^{2} + 4^{2}

9 = PA^{2}

PA = 3

PQ = PA + AQ = 3 + 3 = 6cm

**Q****1****6. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.**

**Ans. **Given : Centre of circle O, PA and PB are tangents drawn at the ends A and B on chord AB.

Prove : ∠PAB = ∠ PBA

Proof : Join OA and OB.

OA = OB (Radii of the same circle)

∠OAB = ∠OBA (Angles opposite to equal sides) —– (i)

∠OAP = ∠OBP = 90 (As radius is always perpendicular to tangent)

∠OAB + ∠PAB = ∠OBA + ∠PBA — (ii)

Putting value of eq (i) in eq (ii) we get,

∠OAB + ∠PAB = ∠OAB + ∠PBA

∠PAB = ∠PBA

Hence proved.

Also Read |
Class 10 Math NCERT Solution |

Also Read |
Class 10 Important Questions [Latest] |