CBSE Solutions For Class 10 Mathematics Chapter 10 Circles

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

⇒  Op² = OT²+ PT²

⇒  (10)² = (8)² + PT²

⇒ 100-64=PT²

⇒  PT² = 36

⇒ PT = 6 Cm

Hence, the length of the tangent is 6cm

Read and Learn More Class 10 Maths

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,

⇒  OA²= op² +Ap²

⇒  (2+r)² = r²+(4)²

⇒  4+r²+4r= r²+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,

⇒  OQ² = Op²= PQ²

= 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