Tangent to a circle: A line which intersects a circle at any one point is called the tangent.

- There is only one tangent at a point of the circle.
- The tangent to a circle is perpendicular to the radius through the point of contact.
- The lengths of the two tangents from an external point to a circle are equal.

## THEOREM 1:

### The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Construction: Draw a circle with centre O. Draw a tangent XY which touches point P at the circle.

To Prove: OP is perpendicular to XY.

Draw a point Q on XY; other than O and join OQ. Here OQ is longer than the radius OP.

OQ > OP

For every point on the line XY other than O, like Q1, Q2, Q3, ……….Qn;

OQ

_{1}> OP
OQ

_{2}> OP
OQ

_{3}> OP
OQ

_{n}> OP
Since OP is the shortest line

Hence, OP ⊥ XY proved

#### THEOREM 2:

##### The lengths of tangents drawn from an external point to a circle are equal.

Construction: Draw a circle with centre O. From a point P outside the circle, draw two tangents P and R.

To Prove: PQ = PR

Proof:

In Δ POQ and Δ POR

OQ = OR (radii)

PO = PO (common side)

∠PQO = ∠PRO (Right angle)

Hence; Δ POQ ≈ Δ POR proved

## Soution of NCERT Exercise 10.1

Question: 1 - How many tangents can a circle have?

Answer: Infinitely many

Question: 2 - Fill in the blanks:

- A tangent to a circle intersect it in ………………..points (s).

Answer: Infinitely many - A line intersecting a circle in two points is called a …………………

Answer: Secant - A circle can have ………….parallel tangents at the most.

Answer: Two - The common point of a tangent to a circle and the circle is called ………………..

Answer: Point of contact

Question: 3 - A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length of PQ is:

## Soution of NCERT Exercise 10.2

Exercise 10.2 - Part - 1

Question: 1 - From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. the radius of the circle is :

- 7 cm
- 12 cm
- 15 cm
- 24.5 cm

Answer: (a) 7 cm

Explanation: Here; PQ = 24 cm, OQ = 25 cm, OP = ?

In ΔOPQ

Question: 2 - In the given figure; if TP and TQ are two tangents to a circle with centre O so that∠POQ=110°, then ∠PTQ is equal to

- 60°
- 70°
- 80°
- 90°

Answer: (b) 70°

Explanation: Here; ∠OPT = ∠OQT = 90° (Since radius is perpendicular to tangent)

Hence, ∠POQ + ∠PTQ = 180°

Or, 110° + ∠PTQ = 180°

Or, ∠PTQ = 180° - 110°

Or, ∠PTQ = 70°

Explanation: Here; ∠OPT = ∠OQT = 90° (Since radius is perpendicular to tangent)

Hence, ∠POQ + ∠PTQ = 180°

Or, 110° + ∠PTQ = 180°

Or, ∠PTQ = 180° - 110°

Or, ∠PTQ = 70°

Question: 3 - If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then POA is equal to

- 50°
- 60°
- 70°
- 80°

Answer: (a) 50°

Explanation: Here; ∠APB = 80°

In ΔPOA;

∠OPA + ∠OAP + ∠POA = 180°

Or, 40° + 90° + ∠POA = 180°

Or, ∠POA = 180° - 130° = 50°

Explanation: Here; ∠APB = 80°

In ΔPOA;

∠OPA + ∠OAP + ∠POA = 180°

Or, 40° + 90° + ∠POA = 180°

Or, ∠POA = 180° - 130° = 50°

Question: 4 - Prove that the tangents drawn at the ends of a diameter are parallel.

Answer: Construction: Draw a circle with centre O. Draw a diameter AB. Draw tangents MN and OP which respectively touch A and B.

To Prove: MN || OP

∠MAB = ∠PBA = 90° (since radius is perpendicular to tangent

Since alternate angles are equal

Hence; MN || OP proved

To Prove: MN || OP

∠MAB = ∠PBA = 90° (since radius is perpendicular to tangent

Since alternate angles are equal

Hence; MN || OP proved

Question: 6 - The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer: Here; OA = 5 cm, AB = 4 cm, OB = ?

Using Pythagoras Theorem,

In ΔOBA

Using Pythagoras Theorem,

In ΔOBA

Question: 7 - 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.

Answer: Here; OA = 5 cm, OP = 3 cm, AB = ?

In ΔOPA,

Since OP bisects AB

Hence, AP = AB

So, AB = 2 x 4 = 8 cm

In ΔOPA,

Since OP bisects AB

Hence, AP = AB

So, AB = 2 x 4 = 8 cm

Question: 8 - A quadrilateral ABCD is drawn to circumscribe a circle. Prove that: AB + CD = AD + BC

Answer: Construction: Draw a circle with centre O. Draw a quadrilateral ABCD which touches the circle at P, Q, R and S.

To Prove: AB + CD = AD + BC

AP = AS

BP = BQ

CQ = CR

DR = DS

(Tangents from same external point are equal)

Adding all the four equations from above; we get:

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

Or, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

Or, AB + CD = AD + BC proved

To Prove: AB + CD = AD + BC

AP = AS

BP = BQ

CQ = CR

DR = DS

(Tangents from same external point are equal)

Adding all the four equations from above; we get:

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

Or, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

Or, AB + CD = AD + BC proved

Question: 9 - In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that AOB = 90°

Answer:

In ΔAPO and ΔACO

AP = AC (tangents from a point)

OP = OC (radii)

OA = OA (common side)

Hence; ΔAPO ≈ ΔACO

So, ∠PAO = ∠CAO

Hence; AO is the bisector of ∠PAC.

Similarly, it can be proved that

BO is the bisector of ∠QBC

Now, XY || X’Y’ (given)

So, ∠AOB = Right angle

(Bisectors of internal angles on one side of transversal intersect at right angle.

In ΔAPO and ΔACO

AP = AC (tangents from a point)

OP = OC (radii)

OA = OA (common side)

Hence; ΔAPO ≈ ΔACO

So, ∠PAO = ∠CAO

Hence; AO is the bisector of ∠PAC.

Similarly, it can be proved that

BO is the bisector of ∠QBC

Now, XY || X’Y’ (given)

So, ∠AOB = Right angle

(Bisectors of internal angles on one side of transversal intersect at right angle.

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.

Answer: Construction: Draw a circle with centre O. Tangent PA and PB are drawn to circle.

To Prove: ∠APB + ∠AOB = 180°

∠OAP = ∠OBP = 90°

In quadrilateral OAPB

∠AOB + ∠APB + ∠OAP + ∠OBP = 360° (sum of all angles of a quadrilateral)

Or, ∠AOB + ∠APB + 90° + 90° = 360°

Or ∠AOB + ∠APB = 360° - 180° = 180° proved

To Prove: ∠APB + ∠AOB = 180°

∠OAP = ∠OBP = 90°

In quadrilateral OAPB

∠AOB + ∠APB + ∠OAP + ∠OBP = 360° (sum of all angles of a quadrilateral)

Or, ∠AOB + ∠APB + 90° + 90° = 360°

Or ∠AOB + ∠APB = 360° - 180° = 180° proved

Question: 11 - Prove that the parallelogram circumscribing a circle is a rhombus.

Answer: Construction: Draw a circle with centre O. Draw a parallelogram ABCD which touches the circle at P, Q, R and S.

Given; AB || DC

AD || BC

To Prove: ABCD is a rhombus

In ΔAOB and ΔDOC;

AB = DC (given that AD || BC)

∠AOB = ∠DOC (Vertically opposite angles)

∠BAO = ∠DCO (Alternate angles)

Hence; ΔAOB ≈ ΔDOC

Hence; AO = CO and BO = DO

Since diagonals are bisecting each other, so given parallelogram is a rhombus.

Given; AB || DC

AD || BC

To Prove: ABCD is a rhombus

In ΔAOB and ΔDOC;

AB = DC (given that AD || BC)

∠AOB = ∠DOC (Vertically opposite angles)

∠BAO = ∠DCO (Alternate angles)

Hence; ΔAOB ≈ ΔDOC

Hence; AO = CO and BO = DO

Since diagonals are bisecting each other, so given parallelogram is a rhombus.

Question: 12 - A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Answer: We have; CD = 6 cm and DB = 8 cm and we need to find AC and AB

For this, we need to use the area of triangle ABC (using Heron’s Formula) and equate the area with sum of areas of triangles AOB, COD and COA. This would help in finding the value of AE or AF.

CD = CE = 6 cm

DB = BF = 8 cm

AE = AF = x

(These are tangents respectively from points C, B and A.

Perimeter 2S = AB + AC + BC = x + 8 + x + 6 + 14 = 2x + 28

Or, S = x + 14

As per Heron’s formula, Area of triangle

Now; area of following triangles can be given as follows:

From equations (1) and (2), we get:

Discarding the negative value; we have AE = 7 cm.

Hence; AB = 7 + 8 = 15 cm

And AC = 7 + 6 = 13 cm

For this, we need to use the area of triangle ABC (using Heron’s Formula) and equate the area with sum of areas of triangles AOB, COD and COA. This would help in finding the value of AE or AF.

CD = CE = 6 cm

DB = BF = 8 cm

AE = AF = x

(These are tangents respectively from points C, B and A.

Perimeter 2S = AB + AC + BC = x + 8 + x + 6 + 14 = 2x + 28

Or, S = x + 14

As per Heron’s formula, Area of triangle

Now; area of following triangles can be given as follows:

From equations (1) and (2), we get:

Discarding the negative value; we have AE = 7 cm.

Hence; AB = 7 + 8 = 15 cm

And AC = 7 + 6 = 13 cm

Question: 13 - Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer: Construction: Draw a circle with centre O. Draw a quadrilateral ABCD which touches the circle at four points.

To Prove: ∠AOB + ∠DOC = 180°

In Δ AOS and ΔAOP;

AS = AP (tangents from A)

AO = AO (common side)

OS = OP (radii)

Hence; Δ AOS ≈ ΔAOP

Hence; ∠1 = ∠8

Similarly, following can be proved:

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

Now;

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

Or, (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°

Or, 2(∠1) + 2(∠2) + 2(∠5) + 2(∠6) = 360°

Or, ∠1 + ∠2 + ∠5 + ∠6 = 180°

Or, (∠1 + ∠2) + (∠5 + ∠6) = 180°

Or, ∠AOB + ∠DOC = 180° Proved

Similarly, ∠AOD + ∠BOC can be proved

To Prove: ∠AOB + ∠DOC = 180°

In Δ AOS and ΔAOP;

AS = AP (tangents from A)

AO = AO (common side)

OS = OP (radii)

Hence; Δ AOS ≈ ΔAOP

Hence; ∠1 = ∠8

Similarly, following can be proved:

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

Now;

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

Or, (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°

Or, 2(∠1) + 2(∠2) + 2(∠5) + 2(∠6) = 360°

Or, ∠1 + ∠2 + ∠5 + ∠6 = 180°

Or, (∠1 + ∠2) + (∠5 + ∠6) = 180°

Or, ∠AOB + ∠DOC = 180° Proved

Similarly, ∠AOD + ∠BOC can be proved

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