WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle Secant To A Circle

A straight line may or may not intersect a circle.

If the straight line does not intersect the circle, then we say that there is no common point between the straight line and the circle.

AB is a straight line and PQRS is a circle which have no common point.

So here the straight line AB does not intersect the circle with centre at O.

Again, if the straight line intersect the circle then we shall see that the straight line can intersect the circle atmost two points.

For example the straight line AB intersects the circle with centre.

O in two points P and Q. In this case, we call the line AB a secant of the circle and PQ is a chord of the secant AB.

It is very clear that the secant AB has intersected the circle with centre O atmost two points P and Q.

So, any secant can intersect a circle atmost two points.

Now, let the straight line AB is rotated anti-clockwise fixing it at the point A.

Where B₁ and B₂ are the new positions of B and in both the positions of B₁ and B₂, the straight line intersects the circle at two points.

Clearly, the new position of P are P₁ and P₂. But in position B₃, P coincides entirely on the point Q.

In this position, the straight line intersects at one point In such cases, P coincides entirely with Q.

In this position the straight line intersect the circle with centre at O only at one point Q,

i.e., there is only one common point between straight line and the circle, which is nothing but Q.

In such a case, we say that the straight line touches the circle with centre at O.

AB is the tangent and Q is the point of contact.

Tangent and point of contact 

Definition: If a straight line intersects a circle in two coincident points, then the straight line is called the tangent to the circle and the point at which they coincide is called the point of contact.

AB is the secant, B₃Q is the tangent and Q is the point of contact. PQ is the corresponding chord of the secant AB.

What do we mean by the term, “Two circles touch each other ?”

Two circles touch each other means

Just before we have seen that two circles can intersect each other atmost in two points.

By joining these two points of intersection, we may get a straight line which is the common chord of two intersecting circles.

Now, if the two centres of the circles be taken away continuously from each other, the length of the common chord gradually decreases and in a certain time two end-points of the common chord coincide at a point on both the circles.

In this position, we say that the two circles touch each other.

For example, PQ is the common chord of the two circles with centres at A and B, when two circles are taken away from each other, then after some time P and Q coincide.

In this position, we say that the two circle touch each other.

 

The two endpoints P and Q of the common chord coincide at point P. In this position, we say that the two circles touch each other externally.

However, if the two circles are derived in one another continuously, then at a certain time, one circle will penetrate completely into another circle, but they remain in contact into an endpoint which is common to both circles.

In this situation we say that the two circles touch each other internally. The two circles with centres at A and B intersect other internally at point C.

Characteristics and properties of two circles touching internally or externally each other 

1. If two circles touch externally, then the distance between centres of the circles is equal to the sum of the radii of the circles.
2. If two circles touch internally, then the distance between the centres of the circles is equal to the difference of their radii.
3. In both cases, the number of point of contact is always 1 and it lie on the same line joining the two centres of the circle.
4. All other points except the point of contact on the tangent to the circle will always lie outside the circle.

Condition for two circles to touch each other 

Let two circles with centres at O and O’ are of radii R₁ and R₂ (R₁ > R₂) respectively and the distance between the centres OO’ = d. Then

1. The two circles will touch each other externally, if R₁ + R₂= d.

2. The two circles will touch each other internally if R₁ – R₂ = d (when R₁ > R₂) and R₂ – R₁ = d when (R₂ > R₁).

3. The-two circles will not touch each other if R₁ – R₂ < d.

 

Common tangent 

Definition: If a straight line touches two given circles at one or two points, then the straight line is called the common tangent to the circles.

For example, the straight line PQ touches two circles with centres at A and B at the point C.

So, here PQ is a common tangent to the circles. Notice that here the common tangent touches the circles at only one point.

Again, the common tangent PQ touches the circle with centre A at C₁ and the circle with centre B at C₂. Notice that here the common tangent PQ touches two circles at two different points.

Types of common tangent

Common tangents are of two types. Such as –

1. Direct common tangent ;
2. Transversed common tangent ;

1. Direct common tangent If the points of contact of the common tangent to two given circles lie on the line segment joining the two centres of the circles or lie on the same side of that line, then the common tangent is called direct common tangent. PQ is a direct common tangent to the circles with centres at A and B.

2. Transversed common tangent If the points of contact of the common tangent of two given circles lie on the opposite sides of the line segment joining the centres of the circles, then the common tangent is called the transverse common tangent.

Both PQ and CD are transversed common tangents to the two given circles with centres at A and B.

Relation between tangent at any point of a circle and the radius passing through that point of contact 

Let PT is a tangent to the circle with centre O at the point P and OP is a radius of the circle passing through P.

We shall now try to find out the relation between PT and OP.

Let OP₁, OP₂, OP₃, OP₄, OP, etc. are the distances from O to some points P₁, P₂, P₃, P₄, P on the tangent PT to the circle with centre O.

If we take measures of OP₃, OP₂, OP₃, OP₄, OP etc with the help of a scale, we shall see that among these distances OP is the smallest. So, OP is perpendicular to PT.

Therefore, tangent to a circle and the radius through the point of contact of the tangent are perpendicular to each other. We shall now prove this theorem logically by the method of geometry.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Secant To A Circle Theorems

“Theorems related to tangents in a circle for Class 10 Maths”

Theorem 1. The tangent to a circle at any point on it is perpendicular to the radius passing through the point of contact.

 

Given: AB is a tangent at P to the circle with the centre at O and OP is a radius of the circle through P.

To prove: OP and AB are perpendicular to each other, i.e., OP ⊥ AB.

Construction: Let us take another point Q on the tangent AB and join the points O, Q.

Proof: Any point on the tangent AB except P lie on the outside of the circle. So, OQ must intersects the circle at any point. Let the point of intersection be T.

∴ OT < OQ [T lies between O and Q]

Again, OT = OP.[radii of same circle] OP < OQ.

∴ OP < OQ

Q is a point on the tangent AB, so OP is the smallest among all the straight lines that can be drawn from O, the centre of the circle, to the tangent AB.

We know that the smallest distance is the perpendicular distance.

∴ OP ⊥ AB. (Proved)

Obviously, there arises a question that whether the converse of this theorem is true or not. We shall verify this logically.

Converse Of Theorem 2. The perpendicular drawn on the radius at the endpoint of the radius of a circle will be a tangent to the circle at the endpoint of the radius. 

Given: CD is a radius of the circle with centre at C and AB is perpendicular to CD at the point D.

To prove: The straight line AB is a tangent to the circle with the centre at C at the point D.

Proof: Let AB is not a tangent to the circle with centre at C at the  point D. Then let us construct another tangent RS at the point D to the R circle with centre at C.

Now, since RS is a tangent at the point D of the circle with the centre at C.

∴ CD ⊥ RS

∴ ∠CDS = 90°, but ∠CDB = 90° (Given) [CD ⊥ AB]

∴ ∠CDS = ∠CDB.

i.e., RS will coincide with the straight line AB.

RS is not a tangent at D on the circle with centre C.

Hence AB is a tangent at the point D of the circle with centre C. (Proved)

Corollary: 1. The point at which any radius of a circle intersects the circumference of the circle, if a perpendicular is drawn on the radius at that point, then the perpendicular is a tangent to the circle at that point.

Corollary: 2. Only one tangent can be drawn at any point on the circumference of a circle.

Corollary: 3. The perpendicular drawn to a tangent at the point of contact passes through the centre of the circle.

In the following examples, applications of the above theorem are discussed thoroughly.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Secant To A Circle Examples

“Chapter 4 tangents and circles exercises WBBSE solutions”

Example 1. Prove that the two tangents at the extremities of a diameter of any circle are parallel to each other.

Solution:

Given: Let AB be the diameter of the circle with the centre at O. AC and BD are two tangents at the end-points A and B of AB.

To prove AC | | BD.

Proof: In the circle with the centre at O, AC is a tangent at A and OA is a radius through the point of contact.

∴ OA is perpendicular to AC.

∴ ∠OAC = 1 right angle……… (1)

Again, in the circle with centre at O, BD is a tangent at the point B and OB is a radius through point of contact.

∴ OB ⊥ BD

∴ ∠OBD = 1 right angle……(2)

Now AB has intersected two line segments AC and BD and two adjacent angles on the same side of the transversal AB are ∠BAC and ∠ABD.

Now, ∠BAC + ∠ABD = ∠OAC + ∠OBD

= 1 right angle + 1 right angle [from (1) and (2)]

= 2 right angles

∴ AC and BD are paralled to each other, i.e., AC | | BD. (Proved)

Example 2.  Manas has drawn a circle with centre O of which AB is a chord. A tangent is drawn at the point B which intersects extended AO at the point T. If ∠BAT = 21°, find the value of ∠BTA.

Solution:

Given that

Manas has drawn a circle with centre O of which AB is a chord. A tangent is drawn at the point B which intersects extended AO at the point T. If ∠BAT = 21°

∠BAT = 21°

∴ ∠OBA = 21° [OA = OB, radii of same circle]

Again, BT is a tangent at B of the circle with centre O and OB is a radius passing through B.

∴ OB ⊥ BT, ∠OBT = 90°

Now, ∠ABT = ∠ABO + ∠OBT

= 21° + 90° [∠ABO= ∠OBA – 21° and ∠OBT =90°] = 111°

Then, ∠BTA = 180° – (∠BAT + ∠ABT) [sum of three angles of a triangle is 180°]

= 180° – (21° + 111°) = 180° – 132° – 48°

Hence ∠BTA = 48°.

Example 3. XY is a diameter of a circle. PAQ is a tangent to the circle at the point A lying on the circumference. The perpendicular drawn on the tangent to the circle from X intersects PAQ at Z. Prove that XA is a bisector of ∠YXZ.

Solution:

Given:

XY is a diameter of a circle. PAQ is a tangent to the circle at the point A lying on the circumference. The perpendicular drawn on the tangent to the circle from X intersects PAQ at Z.

Let XY be a diameter of the circle with centre at O. PAQ is a tangent to the circle at point A. The perpendicular XZ drawn from X to PAQ intersects
PAQ at Z.

To prove: XA is a bisector of ∠YXZ, i.e., ∠YXA = ∠ZXA

Construction: Let us join O, A.

Proof: PAQ is a tangent to the circle with centre O at the point A and OA is a radius passing through point of contact A.

∴ OA ⊥ PAQ

∴ ∠OAP = 90° or, ∠OAZ = 90°…….. (1)

Again, given that XZ ∠ PAQ,

∴ ∠XZA = 90° …. (2)

From (1) and (2) we get, XZ | | OA and XA is their transversal.

∴ ∠OAX = ∠ZXA…….(3)

or, ∠ZXA = ∠AXO [OX = OA = radii of same circle]

or, ∠ZXA = ∠YXA

Hence XA is the bisector of ∠YXZ(proved)

“Class 10 Maths properties of tangents to a circle”

Example 4. PR is a diameter of a circle. A tangent is drawn at the point P and a point S is taken on the tangent of the circle in such a way that PR = PS. If RS intersects the circle at the point T., prove that ST = PT. 

Solution:

Given:

PR is a diameter of a circle. A tangent is drawn at the point P and a point S is taken on the tangent of the circle in such a way that PR = PS. If RS intersects the circle at the point T.,

Let PR be a diameter of the circle with centre at O. A tangent AB is drawn at P.

The part PS from.AB is cut equal to PR. RS intersects the circle at the point T.

To prove: ST = PT

Proof: AB is a tangent at P on the circle with centre at O and OP is a radius passing through P.

∴ OP ⊥ AB.

∴ ∠OPS = 90° or, ∠RPT + ∠TPS = 90° ……… (1)

Again, ∠PTR is a semicircular angle, ∠PTR = 90°

∴ ∠TPR + ∠TRP = 90°…….(2)

From (1) and (2) we get, ∠RPT + ∠TPS = ∠TPR + ∠TRP

or, ∠TPS = ∠TRP

∠RPT = ∠TPR or, ∠TPS = ∠TSP [PS = PR, ∴ ∠TRP = ∠TSP]

∴ ST = PT [opposite sides of equal angles are equal.]

Hence ST = PT (Proved).

Example 5. Two radii OA and OB of a circle with centre O are perpendicular to each other. If two tangents are drawn at the points A and B intersect each other at the point T, prove that AB = OT and they bisect each other at a right angle. 

Solution:

Given:

Two radii OA and OB of a circle with centre O are perpendicular to each other. If two tangents are drawn at the points A and B intersect each other at the point T

Two radii OA and OB of a circle with centre at O are perpendicular to each other.

Two tangents ST and RT at the points A and B respectively intersect each other at T.

Let us join the points A, B and O, T. Let AB and OT intersect each other at C.

To prove: AB = OT and AB and OT bisect each other at a right angle.

Proof: In the quadrilateral OATB, OA ⊥ OB (given), ∠AOB = 90°, OA ⊥ TS, ∠OAT = 90° and OB ⊥ TR.

∴ ∠OBT = 90°, i.e., three of the angles of OATB are right-angles. So, the fourth angle is also a right angle,

∴ each of the angles of OATB is a right angle.

Again, OA = OB [radii of same circle]

∴ OA = OB = BT = TA.

∴ each of the sides of the quadrilateral OATB are equal and each of the angles is a right angle.

∴ OATB is a square and AB and OT are two of its diagonals.

We know that diagonals of a square are equal and the diagonals bisect each other at right-angles.

∴ AB = OT and AB and OT bisect each other at a right angle. [Proved]

“Understanding tangent theorems in Class 10 Maths”

Example 6. X is a point on the tangent at the point A lies on a circle with centre O. A secant drawn from a point X intersects the circle at the points Y and Z. If P is the mid-point of YZ, prove that XAPO or XAOP is a cyclic quadrilateral.

Solution:

Given:

X is a point on the tangent at the point A lies on a circle with centre O. A secant drawn from a point X intersects the circle at the points Y and Z. If P is the mid-point of YZ

XT is a tangent at A of the circle with centre at O. A secant XZ through X intersects the circle at the point Y and Z. P is the mid-point of YZ.

To prove: XAPO or XAOP is a cyclic quadrilateral.

Proof: YZ is a chord of the circle with centre at O and P is the mid-point of YZ.

∴ OP ⊥ YZ.

∴ ∠OPX = 90°

Again XT is a tangent at A of the circle with centre at O.

∴ OA ⊥ XT, ∠OAX = 90°……. (2)

Now. from (1) and (2) we get,

∠OPX + ∠OAX = 90° + 90° = 180°

∴ Two opposite angles of the quadrilateral XAOP are supplementary.

∴ XAOP is a cyclic quadrilateral. (Proved)

Example 7. P is any point on diameter of a circle with centre O. A perpendicular drawn on diameter at the point O intersects the circle at the point Q. Extended QP intersects the circle at the point R. A tangent drawn at the point R intersects extended OP at the point S. Prove that SP = SR.

Solution:

Given:

P is any point on diameter of a circle with centre O. A perpendicular drawn on diameter at the point O intersects the circle at the point Q. Extended QP intersects the circle at the point R. A tangent drawn at the point R intersects extended OP at the point S.

P is any point on a diameter EF of the circle with .centre O.

OQ ⊥ EF and OQ intersects the circle at the point Q.

Extended QP intersects the circle at the point R.

The tangent TS of the circle at R intersects extended OP at the point S.

To prove: SP = SR.

Construction: Let us join O, R.

Proof : OR is a radius of the circle passing through the point of contact of the tangent SRT. ∴ OR ⊥ ST.

∴ ∠ORS = 90° or, ∠ORQ + ∠QRS = 90°…….(1)

Again, OQ ⊥ ES, ∠QOS = 90°, ∠OQP + ∠OPQ = 90°……..(2)

from (1) and (2) we get, ∠ORQ + ∠QRS = ∠OQP + ∠OPQ….. (3)

Now, OQ = OR [radii of same circle] ∴ ∠ORQ = ∠OQR

∴ from (3) we get, ∠OQR + ∠QRS = ∠OQP + ∠OPQ [∠ORQ = ∠OQR]

or, ∠OQR + ∠QRS = ∠OQR + ∠SPR [∠OPQ = opposite ∠SPR]

or, ∠QRS = ∠SPR or, ∠PRS = ∠SPR

∴ SP = SR [ in ΔSPR, sides opposite of equal angles are equal] .

∴ SP = SR (Proved)

“Step-by-step solutions for tangents in circles Class 10”

Example 8. QR is a chord of the circle with centre O. Two tangents drawn at the points Q and R intersect each other at point P. If QM is a diameter, prove that ∠QPR = 2 ∠RQM

Solution:

Given:

QR is a chord of the circle with centre O. Two tangents drawn at the points Q and R intersect each other at point P.

QR is a chord of the circle with centre O. Two tangents drawn at the point Q and R intersect each other at the point P. QM is a diameter.

To prove: ∠QPR = 2 ∠RQM

Construction: Let us join O, R.

Proof: OQ1 PS,  ∴ ∠OQP = 90°

Again OR ⊥ PT, ∴ ∠ORP = 90°

In quadrilateral OQPR, ∠QOR + ∠ORP + ∠RPQ + ∠PQO = 360°

or, ∠QOR + 90° + ∠RPQ + 90° = 360°

or, ∠QOR + ∠RPQ = 180° or, ∠RPQ = 180° – ∠QOR

or, ∠RPQ = ∠ROM……. (1) [∠QOR + ∠ROM = 180°]

Now, ∠RQM is the angle in circle and ∠ROM is the central angle produced by the chord RM.

∴ ∠ROM = 2 ∠RQM

or, ∠QPR = 2 ∠RQM [by (1)]

Hence ∠QPR – 2 ∠RQM. (Proved)

Example 9. Two chords AC and BD of a circle intersect each other at point O. If two tangents drawn at the points A and B intersect each other at the point P and two tangents drawn at the points C and D intersect at the point Q, prove that ZP + ZQ = 2

Solution:

Given:

Two chords AC and BD of a circle intersect each other at point O. If two tangents drawn at the points A and B intersect each other at the point P and two tangents drawn at the points C and D intersect at the point Q

In the circle with centre O’, two chords AC and BD intersect each other at O.

Two tangents PAS and PBT drawn at the points A and B respectively intersect each other at P.

Again, the tangents QCN and QDM drawn at C and D respectively intersect each other at Q.

To prove: ∠P + ∠Q = 2 ∠BOC

Construction: Let us join O’, A; O’, D; A, B and C, D.

Proof: From example 8 above, we get P = 2 ∠BAO’ and Q = 2 ∠DCO’

∴ ∠P + ∠Q = 2 (∠BAO’ + 2 ∠DCO’)

= 2 (∠BAO’ + ∠DCO’)

= 2 (∠BAC + ∠CAO’ + ∠DCO’) [∠BAO’ = ∠BAC = ∠CAO’]

= 2 (∠BDC + ∠ACO’ + ∠DCO’)

= 2 (∠BDC + ∠ACD) [∠ACO’ + ∠DCO’ = ∠ACD]

= 2 (∠ODC + ∠OCD) [∠BDC = ∠ODC, ∠ACD = ∠OCD]

= 2 ∠BOC [in ΔCOD external ∠BOC = internally opposite (∠ODC + ∠OCD]

Hence ∠P +∠Q = 2 ∠BOC (Proved)

Example 10. Prove that if a quadrilateral is circumscribed about a circle, then the angles subtended at the centre by any two opposite sides are supplementary. 

Solution:

Let the quadrilateral ABCD is circumscribed about the circle with centre O.

Two of its opposite sides AB and CD have subtended the angles ∠AOB and ∠COD at the centre.

To prove: ∠AOB and ∠COD are supplementary to ∠AOB + ∠COD = 180°

Construction: Let the sides AB, BC, CD and DA of the quadrilateral ABCD touch the circle at the points P, Q, R and S respectively.

Let us join the points O, A; O, P; O, B; O, Q; O, C; O, R; O, D and O, S.

Proof: AB is a tangent to the circle with centre O and OP is a radius passing through the point of contact P.

OP ⊥ AB. ∠OPA = ∠OPB = 90°

Similarly, OQ ⊥ BC, OR ⊥ CD, OS ⊥ AD.

∴ ∠OQB = ∠OQC = 90°, ∠ORC = ∠ORD = 90°, ∠OSD = ∠OSA = 90°.

Again, PA || OS, [∠OPA = ∠OSA = 90°] and OA is their transversal,

∴ ∠OAP = alternate ∠AOS.

Similarly, ∠OBP = alternate ∠BOQ; ∠COR = alternate ∠OCQ.

∠DOR = alternate ∠ODS; ∠DOS = alternate ∠ODR.

Now, ∠AOB + ∠COD = ∠AOP + ∠BOP + ∠COR + ∠DOR

= ∠AOS + ∠BOQ + ∠COQ + ∠DOS

= ∠AOS + ∠DOS + ∠BOQ + ∠COQ = ∠AOD + ∠BOC

= 360° – (∠AOB + ∠COD)

or, 2 (∠AOB + ∠COD) = 360°.

or, ∠AOB +∠COD = 180°

Hence ∠AOB + ∠COD = 180° (Proved)

Example 11. PQ is the diameter of the circle with centre O. The tangent drawn at any point R on the circle intersects the tangents drawn at P and Q at two points A and B respectively. Prove that ∠AOB = 1 right angle.

Solution:

Given:

PQ is the diameter of the circle with centre O. The tangent drawn at any point R on the circle intersects the tangents drawn at P and Q at two points A and B respectively.

Let PQ.is a diameter of the circle with centre O.

EF and CD are two of its tangents drawn at P and Q respectively.

R lies on the circle and the tangent AB drawn at R intersects the previous tangents at A and B respectively. Let us join O, A; O, B.

To prove ∠AOB = 1 right angle.

Proof: In ΔAOP and ΔAOR,

∠APO =  ∠ARO [each is right angle]

OR = OP [radii of same circle] and hypotenuse OA is common to both.

∴ ΔAOP ≅ ΔAOR [by the RHS condition of congruency]

∴ ∠OAP = ∠OAR ……(1) [similar angles of two congruent triangles]

Similarly, it can be proved that ∠OBR = ∠OBQ…….(2)

Then, ∠AOB = ∠AOR + ∠BOR

= 90° – ∠OAR + 90° – ∠OBR

= 90° – ∠OAP + 90° – ∠OBQ

= ∠AOP + ∠BOQ

= 180° – ∠AOB [∠AOB + ∠AOP + ∠BOQ = 180°] or, 2 ∠AOB = 180° or, ∠AOB = 90°

Hence ∠AOB = 1 right angle. (Proved)

Example 12. The length of radii of two circles are r₁ unit and r₂ unit respectively where r₁ > r₂. If the distance between the centres of the circles be p unit, then prove that the length of the direct common tangent to -the two circles, PQ = \(P Q=\sqrt{p^2-\left(r_1-r_2\right)^2} \text { unit. }\) unit.

Solution:

Given:

The length of radii of two circles are r₁ unit and r₂ unit respectively where r₁ > r₂. If the distance between the centres of the circles be p unit

Let the length of the radius of the circle with centre C₁ be r₁ unit and the length of the radius of the circle with centre C₂ be r₂ unit. (r₁ > r₂).

RS is a direct common tangent to both the circles which intersects the circle with centre C₁ at P and the circle with centre C₂ at Q respectively.

To prove PQ = \(P Q=\sqrt{p^2-\left(r_1-r_2\right)^2} \text { unit. }\)unit,, where p = distance between the centres of the circles;

Construction: Let us draw a perpendicular C₂M from C₂ to PC₁. Let us join P, C₂.

Then PM = QC₂ = r₂

∴ MC₁ = PC₁ – PM

= r₁ – QC₂

= r₁ – r₂

Again, PQ = MC₂

Proof: In the right-angled ΔMC₁C₂ [∠C₁MC₂ = 90°],

MC₁² + MC₂² = C₁C₂² [by Pythagoras’ theorem]

or, (r₁ – >₂)² + PQ² = p² [MC₁ = 1- r₁ – r₂, MC₂ = PQ and C₁C₂ = p]

or, PQ² = p² – (r₁ – r₂)² or, PQ = \(P Q=\sqrt{p^2-\left(r_1-r_2\right)^2} \text { unit. }\)

Hence PQ = \(P Q=\sqrt{p^2-\left(r_1-r_2\right)^2} \text { unit. }\)  unit. (Proved)

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle

The number of tangents to a circle from an external point

In the definition of tangent, we have said earlier that if any straight line intersect a circle at only one point, then the straight line is called a tangent of the circle.

Now, from an external point of a circle, two tangents can be drawn at two points, one point lying above the centre and the other point lying below the centre of the circle.

Such as, let T be an external point of a circle with centre at O. TP and TQ are two tangents drawn from T at the points P and Q respectively on the circle.

Later on, we shall prove it logically.

Number of direct common tangents to a circle 

There can be drawn at most three direct common tangents to two given circles.

Number of transverse common tangents to a circle 

There can be atmost two transverse common tangents to two given circles.

Now, let T be an external point of a circle with centre at O.

Two tangents TP and TQ are drawn from T on the points P and Q on the circle. ∠POT and ∠QOT are two front central angles produced by the tangents TP and TQ.

If we take measures of TP and TQ and the angles ∠POT and ∠QOT with the help of a scale and a protractor, then we shall see that TP and TQ are equal in length and ∠POT = ∠QOT,

i.e., the lengths of the tangents are equal and they produce equal front angles at the centre.

We shall now prove this theorem logically by the method of geometry.

 

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Theorems

“WBBSE Mensuration Chapter 4 practice questions on tangents”

Theorem 1. If two tangents are drawn to a circle from a point outside it, then the line segments joining the point of contact and the exterior point are equal and they subtend equal angles at the centre.

Given: Let PA and PB be two tangents drawn from an external point P of the circle with centre at O and let A and B be their points of contact.

After joining O, A; O, B and O, P, the two line segments PA and PB subtends two angles ∠POA and ∠POB at the centre.

To prove (1) PA = PB and (2) ∠POA = ∠POB.

Proof: PA and PB are two tangents and OA and OB are two radii passing through points of contact.

∴ OA ⊥ PA and OB ⊥ PB .

∠OAP = 90° and ∠OBP = 90°

Now, in the right-angled triangles ΔAOP and ΔBOP, ∠OAP = ∠OBP, [each is a right angle]

OA = OB [radii of same circle] and hypotenuse OP is common to both.

∴ ΔAOP ≅ ΔBOP [by the RHS condition of congruency]

∴  PA = PB [similar sides of congruent triangles] and ∠POA = ∠POB [similar sides of congruent triangles]

Hence PA = PB [(1) is proved]

and ∠POA = ∠POB [(2) is proved]

we have seen that two circles can touch each other in two ways-internally or externally.

In either cases, the point of contact of the two circles always lie on the line segment joining the two centres of the two circles.

We shall now prove this theorem logically by the method of geometry.

Theorem 2. If two circles touch each other, then the point of contact will lie on the line segment joining the two centres.

 

Given: Let two circles with centres A and B touch each other at point P.

To prove: A, P and B are collinear.

Construction: Let us join A, P and B, P.

Proof: Two circles with centres at A and B touch each other at point P.

∴ the two circles have a common tangent at P. Let ST be the common tangent which has touched both the circles at P.

Now,since in the circle with centre A, ST is a tangent to the circle and AP is its radius passing through point of contact P. ∴ AP ⊥ BP.

Again, in the circle with centre at B, ST is a tangent and BP is a radius passing through point of contact P, ∴ BP ⊥ ST.

∴ AP and BP are both perpendicular to ST at the same point P.

AP and BP lie on the same straight line, i.e., the points A, P and B are collinear. (Proved)

In the following examples various applications of the above theorems in practical problems have been discussed throughly.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Multiple Choice Questions

Example 1. In a circle with centre O, a tangent is drawn from an external point A to the circle which touches the circle at B. If OB = 5 cm, OA = 13 cm, then the length of AB 

1. 1. 12 cm
2. 13 cm
3. 6.5 cm
4. 6 cm

 

Solution: In the circle with centre O, AB is a tangent and OB is a radius passing through point of contact.

∴ OB ⊥ AB, ∠OBA = 90°

∴ ΔAOB is a right-angled triangle* of which OA is the hypotenuse.

∴ OB² + AB² = OA² [by Pythagoras theorem] or, 5² + AB² =13² [OB = 5 cm, OA = 13 cm]

or, 25 + AB² = 169 or, AB² = 169 – 25

or, AB² = 144 or, AB = √144 or, AB = 12

Hence the length of AB = 12 cm

∴ 1. 12 cm is correct.

Example 2. Two circles touch each other externally at the point C. AB is a common tangent to both the circles and touches the circle at the points A and B. Then the value of ∠ACB is

1. 60°
2. 45°
3. 30°
4. 90°

Solution: Let two circles with centres O and O’ touch each other externally at point C.

Let us join OCO’, O,A; O’,B; A, C and B, C.

Now, ∠OCO’ = 1 straight angle =180°

or, ∠OCA + ∠ACB + ∠O’CB = 180°

or, ∠OCA + ∠ACB + ∠O’BC =180°

[∠OAC = ∠OAC, ∠O’CB = ∠O’BC]

or, 90° – ∠BAC + 90° – ∠ABC + ∠ACB = 180°

[∠OAC = 90° – ∠BAC,- ∠O’BC = 90° – ∠ABC]

or, 180° – ∠BAC – ∠ABC + ∠ACB = 180°

or, ∠BAC + ∠ACB + ∠ABC – ∠BAC – ∠ABC + ∠ACB = 180°

[sum of three angles of a triangle is 180°.]

or, 2 ∠ACB = 180° or, ∠ACB = \(\frac{180^{\circ}}{2}\)

or, ∠ACB = 90°

∴ 4. 90° is correct.

Example 3. The length of radius of a circle with centre O is 5 cm. P is a point at a distance of 13 cm from the point O. PQ and PR are two tangents from the point P to the circles: the area of the quadrilaterals PQOR is

1. 60 sq-cm
2. 30 sq-cm
3. 120 sq-cm
4. 150 sq-cm

 

Solution: The length of the radius of a circle with centre O. ∴OQ = OR = 5 cm.

P is a point at a distance of 13 cm from the point O. ∴ OP = 13. cm

Now, PQ and PR are two tangents from P and OP and OR are radii passing through the point of contact.

∴ OQ ⊥ PQ and OR ⊥ PR

∴ ∠OQP = 90° and ∠ORP = 90° i.e., ΔPOQ and ΔPOR are both right-angled triangles.

∴ ΔPOQ = \(\frac{1}{2}\) x PQ x OQ [PQ = base and OQ = height]

= \(\frac{1}{2}\) x 12 x 5 sq-cm PQ \(=\sqrt{\mathrm{OP}^2-\mathrm{OQ}^2}=\sqrt{13^2-5^2} \mathrm{~cm}\)

= 30 sq.cm \(\sqrt{169-25} \mathrm{~cm}=\sqrt{144} \mathrm{~cm}=12 \mathrm{~cm}\)

Similarly ΔPOR = \(\frac{1}{2}\) x PR x OR [PR = base and height = OR]

= \(\frac{1}{2}\) x 12 x 5 sq-cm [ PR = PQ = 12 cm] = 30 sq-cm

∴ Area of the quadrilateral PQOR = ΔPOQ + ΔPOR = 30 sq-cm + 30 sq-cm = 60 sq-cm.

∴ 1. 60 sq-cm is correct.

Example 4. The lengths of radii of two circles are 5 cm and 3 cm. The two circles touch externally. Then the distance between the centres of the circles is 

1. 2 cm
2. 2.5 cm
3. 1.5 cm
4. 8 cm

 

Solution: Let two circles with centres O and O’ touch each other externally at the point C.

The radius of the. first circle is 5 cm ∴ OC =5cm, and the radius of the second circle is 3 cm,

∴ O’C = 3cm.

Now, the distance between the centres, of the circles = OO’ = OC + CO’ = 5 cm + 3 cm = 8 cm

∴ 4. 8 cm is correct.

Example 5. The lengths of radii of two circles are 3.5 cm and 2 cm. They touch each other internally. the distance between the centres of the circle is

1. 5.5 cm
2. 1 cm
3. 1.5 cm
4. None of thsese

 

Solution: Let two circles with centres O and O’ touch each other internally at the point C

The radii of the circles are 3.5 cm and 2 cm ∴ OC = 3.5 cm, O’C = 2cm.

Now, the distance between the centres of the circles = OO’ = OC – O’C = 3.5 cm – 2 cm = 1.5 cm

∴ 3. 1.5 cm is correct.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle True Or False

Example 1. P is a point inside a circle; No tangent of the circle will pass through P.

Solution: True

Example 2. In a circle more than two tangents can he drawn which are parallel to a fixed straight line.

Solution: False

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Fill In The Blanks

Example 1. If a straight line intersects the circle at two points, then the straight line is called a ________ to the circle. 

Solution: Secant

Example 2. If two circles do not intersect or touch each other, then the maximum number of common tangents can be drawn is _______

Solution: 4

Example 3. Two circles touch each other externally at point A. A common tangent drawn to two circles at the point A is a ________ common tangent, (direct/transverse).

Solution: Transverse

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Short Answer Type Questions

Example 1.  In the adjoining O is the centre and BOA is the diameter of the circle. A tangent drawn to the circle at the point P intersects the extended BA at point T. If ∠PBO = 30°, find the value of ∠PTA.

Solution:

Given:

In the adjoining O is the centre and BOA is the diameter of the circle. A tangent drawn to the circle at the point P intersects the extended BA at point T. If ∠PBO = 30°,

Let us join O, P and A. P.

PT is tangent to the circle at P and OP is a radius through the point of contact P.

∴ OP ⊥ PT

∴ ∠OPT = 90°, ∠OPB – ∠OBP = 30° ∠BPT = ∠OPB + ∠OPT

= 30° + 90° [∠OPB = 30° and ∠OPT = 90°] = 120°

Now, in ΔBPT, ∠BTP + ∠TPB + ∠PBT = 180° or, ∠BTP + 120° + 30° = 180°

or, ∠PTA + 150° = 180° or, ∠PTA = 180° – 150° or, ∠PTA = 30°

∴ The value of ∠PTA = 30°.

Example 2. In the adjoining ΔABC circumscribes a circle and touches the circle at the points P. Q, R. If AP = 4 cm, BP = 6cm, AC = 12 cm and BC = x cm, then determine the value of x. 

Solution:

Given:

In the adjoining ΔABC circumscribes a circle and touches the circle at the points P. Q, R. If AP = 4 cm, BP = 6cm, AC = 12 cm and BC = x cm

Let us join O, P; O, Q; O,A, O,B and O,C.

Now ΔAOP = ΔAOR [OP = OR, ∠OPA = ∠ORA = 90° and hypotenuse OA is common to both.]

∴ AP = AR [similar sides of congruent triangles]

Then CR = AC – AR

= 12 cm – AP [AR = AP]

= 12 cm – 4 cm [AC = 12 cm and AP = 4 cm] = 8 cm

Again, ΔCOQ ≅ ΔCOR [by the R-H-S condition of congruency]

CQ = CR [similar sides of congruent triangles]

= 8 cm [CR = 8 cm]…….(1)

Again, ΔBOQ ≅ ΔBOP. [by the R-H-S condition of congruency]

∴ BQ = BP [similar sides of congruent triangles]

= 6 cm [BP = 6 cm]……… (2)

Then, BC = BQ + CQ = 6 cm + 8 cm [from (1) and (2)]

∴ BC = x cm = 14 cm;

The value of x = 14 cm.

“Examples of tangent problems for WBBSE Class 10 Maths”

Example 3. In the adjoining three circles with centres A, B, and C touch one another externally. If AB =  5 cm, BC = 7 cm and CA = 6 cm, then find the length of radius of a circle with centre A. 

Solution:

Given:

In the adjoining three circles with centres A, B, and C touch one another externally. If AB =  5 cm, BC = 7 cm and CA = 6 cm

As per the question, AB = 5 cm.

∴ AP + BP = 5 [two circles with centres A and B touch each other externally at P]………(1)

Similarly, BC = 7 cm

∴ CR + BR = 7 cm ……….. (2)

and CA = 6 cm, ∴ AQ + CQ = 6cm

or, AQ + CR = 6 cm…..(3)

[CQ = CR, radii of same circle.]

Then from (2) and (3) we get,

BR – AQ = 1 cm.

or, BP – AP = 1 cm [BR = BP and AQ = AP] ….. (4)

Now, subtracting (4) from (1) we get,

2 AP = 4 cm or, AP = \(\frac{4}{2}\) cm = 2 cm

∴ the length of the radius of the circle with centre A = 2 cm.

Example 4. In the adjoining two tangents drawn from external point C to a circle with centre O touches the circle at the points P and Q respectively. A tangent drawn at another point R of the circle intersects CP and CQ at the points A and B respectively. If CP = 11 cm and BC = 7 cm, then determine the length of BR.

 

Solution:

Given:

In the adjoining two tangents drawn from external point C to a circle with centre O touches the circle at the points P and Q respectively. A tangent drawn at another point R of the circle intersects CP and CQ at the points A and B respectively. If CP = 11 cm and BC = 7 cm,

Let us join O, P ; O, Q ; O, R and O, B.

Now, in the circle with centre O, CP and CQ are two tangents from an external point C.

∴ CP = CQ

∴ CQ = 11cm [CP = 11 cm (given)]

Again, given that BC = 7 cm

∴ BQ = CQ – BC

= 11 cm – 7cm = 4 cm

Now, in ΔBOQ and ΔBOR, OQ = OR, [radii of same circle]

∴ ∠OQB = ∠ORB [each is right angles] and hypotenuse OB is common to both.

∴ ΔBOQ = ΔBOR [by the R-H-S condition of congruency]

∴ BQ = BR [similar sides of congruent triangles]

But BQ = 4 cm, ∴ BR = 4 cm.

The length of BR = 4 cm

Example 5. The length of radii of two circles are 8 cm and 3 cm and distance between two centres is 13 cm. Find the length of a direct common tangent of two circles.

Solution:

Given:

The length of radii of two circles are 8 cm and 3 cm and distance between two centres is 13 cm.

Let the radii of the circles with centres A and B are 8 cm and 3 cm respectively and . the distance PQ is the direct common tangent of both the circles.

We have to find the length of PQ. Let us join A, P and B, Q and let us draw BM perpendicular to AP from B.

Then PQBM is a rectangle. [∠APQ  – ∠BQP = 90°]

∴ PQ = BM and PM = BQ.

As per question, AP = 8 cm, BQ = 3 cm.

Then AM = AP – PM = 8 cm – BQ = 8 cm – 3 cm = 5 cm

Now, in the right-angled triangle ABM we get by Pythagoras theorem,

BM² + AM² = AB² [∴ AB = hypotenuse] .

or, BM² + 5² = 13² or, BM² = 13² – 5² = 169 – 25 = 144

∴ BM = √144 = 12

∴ PQ = 12 cm [BM = PQ]

∴ the required length of the direct common tangent of two circles is 12 cm.

Solid Geometry Chapter 4 Theorems Related To Tangent In A Circle

Tangent To A Circle Long Answer Type Questions

“WBBSE Class 10 Maths solved problems on circle tangents”

Example 1. An external point is situated at a distance of 17 cm from the centre of a circle having 16 cm diameter. Determine the length of the tangent drawn to the circle from the external point.

Solution:

Given:

An external point is situated at a distance of 17 cm from the centre of a circle having 16 cm diameter.

Since the diameter of the circle with centre O is 16 cm.

∴ radius = \(\frac{16}{2}\) cm = 8 cm.

PT is a tangent to the circle from an external point P which is situated at a distance .of 17 cm from the centre of the circle ∴ OP = 17 cm.

We have to determine the length of PT.

Now, PT is a tangent at T to the circle with centre O and OT is the radius passing through point of contact.

∴ OT ⊥ PT, ∴ ∠PTO = 1 right angle.

So, from the right-angled triangle POT by Pythagoras theorem,

OP² = OT² + PT²

or, 17² = 8² + PT² [OP = 17 cm, OT = 8 cm]

or, 289 = 64 + PT² or, PT² = 289 – 64 .

or, PT² = 225 or, PT = V225 = 15.

∴ the required length of the tangent = 15 cm.

Example 2. The tangent drawn at points P and Q on the circumference of a circle intersect at A. If ∠PAQ = 60°, find the value of ∠APQ. 

Solution:

Given:

The tangent drawn at points P and Q on the circumference of a circle intersect at A. If ∠PAQ = 60°,

Let P and Q be any two points on the circle with centre O. Two tangents AP and AQ drawn at the points P and Q intersect each other at point A.

Let us join P, Q and O, A.

Given that ∠PAQ = 60°.

Now, AP is a tangent to the circle with centre O and OP is the radius passing through the point of contact P.

∴ OP ⊥ AP, ∴  ∠OPA = 90°.

Similarly, ∠OQA = 90°

In Δ’s POA and ΔQOA,

∠OPA = ∠OQA [each is a right-angle]

OP = OQ [radius of same circle] and hypotenuse OA is common to both A POA = AQOA [by the R-H-S condition of congruency]

∴ ∠OAP = ∠OAQ [similar angles of congruent triangles]

Given that ∠PAQ = 60° or, ∠PAO + ∠QAO = 60° or, ∠PAO + ∠PAO = 60° [∠QAO = ∠PAO] = 60°

or, 2 ∠PAO = 60° or, ∠PAO = \(\frac{60^{\circ}}{2}\) = 30°

Then, ∠POA = 90° – ∠OAP = 90° – 30° = 60° [∠OAP = ∠PAO = 30°]

∴ ∠POQ = 2 ∠POA [∠POA = ∠QOA] = 2 x 60° = 120°

∴ ∠OPQ + ∠OPQ – 180° – 120° [∠OQP = ∠OPQ and∠POQ = 120°]

or, 2 ∠OPQ = 60° or, ∠OPQ = \(\frac{60^{\circ}}{2}\)= 30°

∴ ∠APQ = ∠APO – ∠OPQ = 90° – 30° – 60°

∴ value of ∠APQ = 60°.

Example 3. AP and AQ are two tangents drawn from an external point A to a circle with centre O, P and Q are points of contact. If PR is a diameter, prove that OA | | RQ. 

Solution:

Given:

AP and AQ are two tangents drawn from an external point A to a circle with centre O, P and Q are points of contact. If PR is a diameter

AP and AQ are two tangents to the circle with centre O and OP and OQ are two radii of the circle passing through points of contact P and Q respectively.

∴ OP ⊥ AP and OQ ⊥ AQ

∴ ∠OPA = ∠OQA = 90°

Now, ΔAOP = ΔAOQ, [since ∠APO = ∠AQO, OP = OQ and hypotenuse OA is common to both.]

∴ ∠AOP = ∠AOQ

[similar angles of congruent triangles]……… (1)

∴ ∠POQ = ∠AOP + ∠AOQ

= ∠AOQ + ∠AOQ [from (1)] = 2 ∠AOQ ……..(2)

Again, ∠POQ is the central angle and ∠PRQ is the angle in circle produced by the arc PQ,

∴ ∠POQ = 2 ∠PRQ

or, 2 ∠AOQ = 2 ∠PRQ [from (2)] .

or, ∠AOQ = ∠PRQ = ∠OQR [OQ = OR, ∴ ∠ORQ = ∠OQR]

∴ ∠AOQ = ∠OQR.

But these are two alternate angles when OQ intersects the line segments OA and RQ, and they are also equal.

∴ OA | | RQ. (Proved)

Example 4. Prove that a parallelogram circumscribing by a circle is always a rhombus. 

Solution:

Let ABCD is a parallelogram circumscribed by a circle with a centre at O.

We have to prove that ABCD is a rhombus.

Proof: Let the circle touches the parallelogram ABCD at the points P, Q, R and S.

Then AP and AS are two tangents to the circle from the external point A.

∴ AP = AS…… (1)

Similarly, BP = BQ …….(2)

CQ = CR ……… (3) and

DR = DS ……. (4)

Now, ABCD is a parallelogram, AB = DC and AD = BC.

Now, AB + DC = AP + BP + DR + CR

= AS + BQ + DS + CQ

= AS + DS + BQ + CQ

= AD + BC

or, AB + AB = BC + BC [DC = AB and AD = BC]

or, 2AB = 2BC or, AB = BC

Similarly, AD = DC, AB = BC = CD = DA

∴ four sides of the parallelogram ABCD are equal and ABCD is also a parallelogram

∴ ABCD is a rhombus. (Proved)

Example 5. Two circles drawn with centres A and B touch each other externally at C, O is a point on the ṭangents drawn at C, OD and OE are tangents drawn to the circles of centres A and B respectively. If ∠COD = 56°, ∠COE = 40°, ∠ACD = x°, and ∠BCE = y°, then prove that Oc = Od = OE and x-y = 8

Solution:

Given:

Two circles drawn with centres A and B touch each other externally at C, O is a point on the ṭangents drawn at C, OD and OE are tangents drawn to the circles of centres A and B respectively. If ∠COD = 56°, ∠COE = 40°, ∠ACD = x°, and ∠BCE = y°,

OC and OD are two tangents at the points C and D on the circle with centre at A from an external point O.

∴ OC = OD …….(1)

Again, OC and OE are two tangents at the points C and E to the circle with centre B from an external point O.

∴ OC = OE……..(2)

Then from (1) and (2) we get, OC = OD = OE

Now in ΔACD, ∠ADC = ∠ACD = x° [AC = AD]

∴ ∠ODC = ∠OCD = 90° – x° [∠ADO = ∠ACO – 90°]

We know that ∠OCD + ∠ODC + ∠COD = 180°

or, 90° – x° + 90° – x° +56° = 180° [∠COD = 56°] or, 2x° = 56° or, x° = \(\frac{56^{\circ}}{2}\) =28°.

∴ x = 28 ……… (3)

Similarly, ∠BEC = ∠BCE = y° [BC = BE]

∴ ∠OCE = ∠OEC = 90° – y° [∠BCO – ∠BEO = 90°]

We know that in ΔOCE, ∠OCE + ∠OEC + ∠COE = 180°

or, 90° -y°+ 90° – y° + 40° = 180° [∠COE = 40°]

or, 2y° = 40° or, y° = \(\frac{40^{\circ}}{2}\) = 20° .

∴ y = 20 …(4)

Then x – y = 28 – 20 = 8 [from (3) and (4)]

Hence OC = OD = OE and x – y = 8 (Proved)

Example 6. Two circles with centres A and B touch each other internally. Another circle touches the larger circle internally at point X and the smaller circle externally at the point Y. If O be the centre of that circle, prove that (AO + BO) is constant.

Solution:

Given:

Two circles with centres A and B touch each other internally. Another circle touches the larger circle internally at point X and the smaller circle externally at the point Y. If O be the centre of that circle,

Let the circle with centre A touches the circle with centre B internally at the point C.

We know that if two circles touches each other either externally or internally, then the centres of the circles and the point of contact lie on the same straight line.

∴ the points A, O, X; A, B, C and B, Y, O lie on the same straight line each.

Now, AO + BO = AO + BY + OY

= AO + OY + BY

= AO + OX + BY [ OY = OX = radii of same circle]

= AX + BY [AO + OX = AX]

= Radius of the larger circle + Radius of the smaller circle.

= Constant [both the circles are fixed]

Hence (AO + BO) = constant (Proved)

“WBBSE Class 10 tangents to a circle solved examples”

Example 7.  Two circles have been drawn with centres A and B which touch each other externally at the point O. A straight line is drawn passing through the point O and intersects the two circles at P and Q respectively. Prove that AP | | BQ.

Solution:

Given:

Two circles have been drawn with centres A and B which touch each other externally at the point O. A straight line is drawn passing through the point O and intersects the two circles at P and Q respectively.

Let two circles with centres A and B touch each other externally at the point O.

∴ A, O, B will lie on the same straight line

Now, in ΔAOP and ΔBOQ, ∠AOP = ∠BOQ [Opposite angles]

Again, in circle with centre at A, AO = AP [radii of same circle]…..(1)

Also, in circle with, the centre at B,

BO = BQ [Radii of same circle]

∠BOQ = ∠BQO …….. (2)

Since ∠AOP = ∠BOQ, from (1) and (2) we get, ∠APO = ∠BQO

But these are the alternate angles produced when two line segments AP and BQ are intersected by the transversal PQ and also they are equal.

∴ AP | | BQ (Proved)

Example 8. Three equal circles touch one another externally. Prove that the centres of the three circles form an equilateral triangle. 

Solution:

Given:

Three equal circles touch one another externally.

Let three equal circles with centres A, B and C touch one another externally at the points P, Q and R.

To prove: ΔABC is an equilateral triangle.

Proof: In the circle with centre A,

AP = AR [radii of same circle]

Similarly, BP = BQ and CQ = CR.

Again, the circles are equal, so the radii of them are all equal, i.e., AP = BP = BQ = CQ = CR = RA…… (1)

Now, AB = AP + BP = BQ + CQ = BC ……(2)

[AP = BQ and BP = CQ, from (1)]

Again, BC = BQ + CQ = CR + AR [BQ = CR and CQ = AR] = CA… (3)

∴ from (2) and (3) we get, AB = BC = CA.

∴ ΔABC is an equilateral triangle. (Proved)

“Tangent properties and theorems for Class 10 students”

Example 9. Two tangents AB and AC drawn from an external point A of a circle touch the circle at the point B and C. A tangent, drawn to a point X lies on minor arc BC intersects AB and AC at the points D and E respectively. Prove that perimeter of ΔADE = 2AB.

Solution:

Given:

Two tangents AB and AC drawn from an external point A of a circle touch the circle at the point B and C. A tangent, drawn to a point X lies on minor arc BC intersects AB and AC at the points D and E respectively.

Let AB and AC are two tangents drawn from an external point A to the circle with centre O at the points B and C respectively.

Similarly, DB = DX and EX = EC……. (2)

Now, perimeter of ΔADE = AD + DE + EA

= AD + (DX + EX) + EA

= AD + (DB + EC) + EA [from (2)]

= (AD + DB) + (AE + EC)

= AB + AC

= AB + AB [AC = AB] = 2AB.

Hence perimeter of ΔADE = 2AB.(Proved)

Example 10. PQ is a diameter. The tangent drawn at the point R, intersects the two tangents drawn at the points P and Q at the points A and B respectively. Prove that ∠AOB is a right angle.

Solution:

Given:

PQ is a diameter. The tangent drawn at the point R, intersects the two tangents drawn at the points P and Q at the points A and B respectively.

Let us join O, R. Also, let OA intersects the arc PR at E and OB intersects the arc RQ at F.

AP = AR, ∴ arc PE = arc RE

The central angles subtended by them are equal, i.e., ∠POE = ∠ROE or, ∠AOP = ∠AOR

Similarly, ∠BOR = ∠BOQ.

Then ∠AOP + ∠AOR + ∠BOR + ∠BOQ = 180° [straight angle]

or, ∠AOR + ∠AOR + ∠BOR + ∠BOR = 180°

[∠AOP = ∠AOR, ∠BOQ = ∠BOR]

or, 2 ∠AOR + 2 ∠BOR = 180° or, 2 (∠AOR + ∠BOR) = 180°

or, 2 ∠AOB = 180° or, ∠AOB = \(\frac{180^{\circ}}{2}\)= 90°

Hence ∠AOB is a right angle. (Proved)