## CBSE Solutions For Class 10 Mathematics Chapter 10 Circles

**Question 1. The radius of a Circle and 8 Cm. Calculate the length of a tangent down to this Circle from a point at a distance of 10 Cm from its Centre. **

**Solution:**

Since the tangent is perpendicular to the radius through the paint of Contact

∠OTP = 90

In the right triangle OTP, we have

⇒ Op^{2} = OT^{2}+ PT^{2}

⇒ (10)^{2} = (8)^{2} + PT^{2}

⇒ 100-64=PT^{2}

⇒ PT^{2} = 36

⇒ PT = 6 Cm

Hence, the length of the tangent is 6cm

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**Question 2. Prove that the tangents drawn at the ends of the diameter of a Circle are Parallel. **

**Solution:**

Let AB be the diameter of a circle with Centre O. PA and PB are the tangents to the Circle at pants A and B respectively.

Now ∠PAB=90°

and∠QBA=90°

⇒ ∠PAB + ∠QBA = 90° +90° = 180°

PA ll QB

**Question 3. prove that the perpendicular at the point of contact to the tangent to a Circler passes through the Centre. **

**Solution:**

**Given:** A Circle with Centre 0 and a PQ tangent AQB and a Perpordicits is a dragon from point of contact Q to AB.

To prove: The perpendicular pa Passes through the Centre of the Circle.

Proof: AQ is the tangent of the Circle at point Q.

AQ will be the perpendicular to the radius of the circle.

⇒ PQ⊥AQ

⇒ The Centre of the Circle will lie on the line PQ.

Perpendicular PQ passes through the Centre of the Circle.

**Question 4. A quadrilateral ABCD is drawn to Circumscribe a circle, and prove that AB+CD = AD+BC. **

**Solution:**

As shown, the sides of a quadrilateral ABCD touch P a the Circle at P, Q, R and s. We know the tangents drawn from an external point to the Clucle are equal.

AP=AS, BP = BQ, CR = CQ, DR = DS

On adding, AP+BP + CR+DR

⇒ AB + BQ + CQ + DS

⇒ AB+CD= (AS + DS) + (BQ+CQ)

⇒ AB + CD = AP+BC

Hence proved.

**Question 5. Ap is tangent to Circle 0 at point P. What is the length of OP? **

**solution:**

Let the radius of the given Circle is r.

OP = OB = r

OA=2+r, OP=r, AP=4

∠OPA = 90°

In the right ∠OPA,

⇒ OA^{2}= op^{2} +Ap^{2}

⇒ (2+r)^{2} = r^{2}+(4)^{2}

⇒ 4+r^{2}+4r= r^{2}+16

⇒ 4r = 12 =) r=3

Op=3cm.

**Question 6. If the angle between two tangents drawn from an external point p to a Clicle of radius ‘a’ and Centre 0, is 60°, then find the length of op .**

**Solution:**

PA and PB are two tangents from an external point p such that

∠APB = 60°

∠OPA = ∠OPB = 30°

(tangents are equally inclined at the centre)

Also, ∠OAP=90°

Now, in right ∠OAP,

Sin 30° =\(=\frac{O A}{O P}\)

⇒ \(\frac{1}{2}=\frac{a}{o p}\)OP=2a units.

**Question 7. In the given figure, if AB = AC, prove that BE = EC. **

**Solution:**

We know that lengths of tangents from an external Point are equal.

AD=AF

DB = BE

EC = FC

Now, it is given that

AB = AC

⇒ AD+DB = AF + EC

⇒ AD+DB = A8+EC

⇒ DB = EC

BE = EC

**Question 8. In the given figure, AT is tangent to the Chicle with Centre 0 Such that Oto 4cm and LOTA = 30° Find the length of Segment AT. **

**Solution:**

In the right ∠OAT,

Cos 30°\(=\frac{A T}{O T}\)

⇒ \( \frac{\sqrt{3}}{2}=\frac{A T}{4}\)

⇒ AT = 2√3 Cm

**Question 9. The length of a tangent from point A at a distance of 5 cm from the Centre P 5cm of the Circle is ucm. Find the radius of the Circle. **

**Solution:**

Let o be the Centre of the Circle and PQ is a tangent to the Circle from point P.

Given that, PQ=4cm and op=5cm

Now,∠OOP = 90°

In ∠OQP,

⇒ OQ^{2} = Op^{2}= PQ^{2}

= 52-42

=25-16=9

OQ = 3cm

Radius of Circle = 3cm

**Question 10. Prove that the angle between the two tangents drawn from an external point to a circle is Supplementary to the angle Subtended by the line segment joining the points of contact at the Centre. **

**Solution:**

PA and PB are the tangents of the Circle.

∠OAP = ∠OBP = 90°

In □ OAPB,

In □ OAPB,

∠OAP + ∠APB +∠OBP + ∠AOB = 360°

⇒ 90°+ ∠APB +90° +∠AOB = 360°

⇒ ∠APB + ∠AOB = 180°

⇒ ∠APB and ∠ADB are Supplementary