In the previous section we saw some basics about tangents. In this section we will learn more details.
Let us do an experiment:
• Consider any point inside a circle. Try to draw a tangent from that point. We can try in any direction we like. We will find that the lines we draw always form a secants. This is shown in fig.32.9(a) below. We will never be able to draw a tangent from a point inside the circle.
• Now take any point on the circle. We have already proved in the previous section that, only one tangent can be drawn through a point on the circle. This is shown in fig(b)
• Now take a point outside the circle. We can try in any direction we like. We will find that only two tangents are possible. This is shown in fig(c)
Case 2: Only one tangent can be drawn through a point lying on the circle
Case 3: Only two tangents can be drawn through a point lying outside the circle
The above details about 'case 3' can be written in the form of a theorem. We will write it in a step by step manner:
2. So there will be two points of contact T1 and T2
3. The lengths PT1 and PT2 will be the same
In the next section, we will see more details about Tangents.
Let us do an experiment:
• Consider any point inside a circle. Try to draw a tangent from that point. We can try in any direction we like. We will find that the lines we draw always form a secants. This is shown in fig.32.9(a) below. We will never be able to draw a tangent from a point inside the circle.
 |
| Fig.32.9 |
• Now take a point outside the circle. We can try in any direction we like. We will find that only two tangents are possible. This is shown in fig(c)
■ So we can write a summary of the three cases shown in the fig.32.9 above.
Case 1: No tangent can be drawn through a point lying inside a circleCase 2: Only one tangent can be drawn through a point lying on the circle
Case 3: Only two tangents can be drawn through a point lying outside the circle
Out of the three cases, 'case 3' requires a detailed discussion. Consider fig.32.10 below:
■ In fig.32.10(a), T1 and T2 are the points of contact.
■ The lengths PT1 and PT2 will be always equal.
■ The measurement from P to T1 (or T2) is called the length of tangent from point P to the circle.
 |
| Fig.32.10 |
■ The lengths PT1 and PT2 will be always equal.
■ The measurement from P to T1 (or T2) is called the length of tangent from point P to the circle.
Theorem 32.2:
1. From any point P outside a circle, two tangents can be drawn2. So there will be two points of contact T1 and T2
3. The lengths PT1 and PT2 will be the same
We can prove the theorem using fig.32.10(b)
1. By applying theorem 32.1, ∠OT1P = 90o. So triangle OT1P is a right angled triangle
2. By applying theorem 32.1, ∠OT2P = 90o. So triangle OT2P is a right angled triangle
3. We will check whether these two triangles ⊿OT1P and ⊿OT2P are congruent or not. Since they are both right angled, we will use the RHS criterion.
(i) In the RHS criterion, we first check the hypotenuse.
• The hypotenuse of ⊿OT1P is OP (∵ OP is opposite the 90o angle)
• The hypotenuse of ⊿OT2P is also OP (∵ OP is opposite the 90o angle)
• So both the triangles have the same hypotenuse
(ii) Next we check the legs
• In the ⊿OT1P, consider the leg OT1. An exact replica of this leg is present in the other ⊿OT2P also.
• This is because OT1 and OT2 are the radii of the same circle.
(iii) So, the two triangles ⊿OT1P and ⊿OT2P:
• Have the same hypotenuse
• Have one leg the same
■ So they are congruent
(iv) We can write the correspondence of corners:
T1 and T2 are the 90o corners. So T1 ↔ T2.
O and P are common corners. So O ↔ O and P ↔ P
(v) From the correspondence of corners, we can write the correspondence of sides:
• OT1 ↔ OT2, OP ↔ OP and PT1 ↔ PT2.
• The last one among the above three is PT1 ↔ PT2. That means, the two sides PT1 and PT2 are equal
4. Thus we proved theorem 32.2
5. We have another useful result also:
(i) Consider the congruence of corners written in 3(iv) above. We have O ↔ O and P ↔ P
• It is clear that the angles at O are equal. That is: ∠T1OP = ∠T2OP
• Also it is clear that angles at P are equal. That is: ∠OPT1 = ∠OPT2
♦ So the angles shown in green and blue colours in fig.32.10(b) are equal
From the results in (5), we can write another theorem. Let us analyse it first:
■ We have: ∠T1OP = ∠T2OP
• But ∠T1OP is the angle subtended by the tangent PT1 at the centre of the circle
• Also ∠T2OP is the angle subtended by the tangent PT2 at the centre of the circle
• The fact that these angles are equal is very helpful for solving problems
■ We have: ∠OPT1 = ∠OPT2
• But ∠OPT1 is the angle that the tangent PT1 makes with the line joining P to the centre O
• Also ∠OPT2 is the angle that the tangent PT2 makes with the line joining P to the centre O
• The fact that these angles are equal is also very helpful for solving problems
We will combine the above results and write it as a theorem. We will write it in steps, and also as two parts:
Theorem 32.3:
1. From any point P outside a circle, two tangents can be drawn
2. So there will be two points of contact T1 and T2.
Part (i):
• Tangent PT1 will subtend ∠T1OP at the centre of the circle
• Tangent PT2 will subtend ∠T2OP at the centre of the circle
• Then ∠T1OP = ∠T2OP
Part (ii):
• A line OP is drawn from the centre O to point P
• Tangent PT1 will make ∠OPT1 with OP
• Tangent PT2 will make ∠OPT2 with OP
• Then ∠OPT1 = ∠OPT2
So we proved theorems 32.2 and 32.3 using the RHS criterion. If we want to prove theorem 32.2 only, we can use an easier method:
We can use the same fig.32.10(b)
1. By applying theorem 32.1, ∠OT1P = 90o. So triangle OT1P is a right angled triangle
2. Using Pythagoras theorem, PT12 = OP2 - OT12
3. By applying theorem 32.1, ∠OT2P = 90o. So triangle OT2P is a right angled triangle
4. Using Pythagoras theorem, PT22 = OP2 - OT22
• But OT2 = OT1.
• This is because OT1 and OT2 are the radii of the same circle.
5. So (4) can be written as: PT22 = OP2 - OT12.
6. But right side of (5) is same as right side of (2)
7. So we can write: PT12 = PT22 ⇒ PT1 = PT2.
Thus theorem 32.2 is proved
Now we will see a solved example
Solved example 32.2:
Two concentric circles C1 and C2 have center at O (fig.32.11.a).
The tangent at a point P on the smaller circle C2, intersects the larger circle at A and B. Prove that AP = BP
Solution:
1. Join P to the center O (fig.32.11.b). Then OP is a radius and it will be perpendicular to AB (Theorem 32.1)
2. Now AB is a chord of C2 and OP is the perpendicular drawn from the center of C2 onto AB
3. Then OP will be bisector of AB (Theorem 17.1).
Thus we get AP = BP
Another method:
1. Join P to the center O (fig.32.11.c). Then OP is a radius and it will be perpendicular to AB (Theorem 32.1)
2. Also join OA and OB. We get two right triangles: ⊿APO and ⊿BPO
3. Next step is to prove that, the two triangles are congruent. Since they are both right angled, we will use the RHS criterion.
Two concentric circles C1 and C2 have center at O (fig.32.11.a).
 |
| Fig.32.11 |
Solution:
1. Join P to the center O (fig.32.11.b). Then OP is a radius and it will be perpendicular to AB (Theorem 32.1)
2. Now AB is a chord of C2 and OP is the perpendicular drawn from the center of C2 onto AB
3. Then OP will be bisector of AB (Theorem 17.1).
Thus we get AP = BP
Another method:
1. Join P to the center O (fig.32.11.c). Then OP is a radius and it will be perpendicular to AB (Theorem 32.1)
2. Also join OA and OB. We get two right triangles: ⊿APO and ⊿BPO
3. Next step is to prove that, the two triangles are congruent. Since they are both right angled, we will use the RHS criterion.
(i) In the RHS criterion, we first check the hypotenuse.
• The hypotenuse of ⊿APO is OA (∵ OA is opposite the 90o angle)
• The hypotenuse of ⊿BPO is also OB (∵ OB is opposite the 90o angle)
• So both the triangles have the same hypotenuse (∵ OA and OB are the radii of C1)
(ii) Next we check the legs
• The leg OP is present in both the triangles.
(iii) So, the two triangles ⊿APO and ⊿BPO:
(iii) So, the two triangles ⊿APO and ⊿BPO:
• Have the same hypotenuse
• Have one leg the same
■ So they are congruent
(iv) We can write the correspondence of corners:
O and P are common corners. So O ↔ O and P ↔ P
A and B are the remaining corners. They will be corresponding corners. So A ↔ B.
A and B are the remaining corners. They will be corresponding corners. So A ↔ B.
(v) From the correspondence of corners, we can write the correspondence of sides:
• OA ↔ OB, OP ↔ OP and AP ↔ BP.
• The last one among the above three is AP ↔ BP. That means, the two sides AP and BP are equal
Solved example 32.3
Two tangents TP and TQ are drawn to a circle with center O from an external point T (fig.32.12.a).
Prove that ∠PTQ = 2 ∠OPQ
Solution:
• Let ∠OPQ = θ (fig.32.12.b). Then we have to prove that ∠PTQ = 2θ
We will write the steps:
1. TP is the tangent drawn from an external point T to a point P on the circle
• OP is the line joining that P to the center O
• So ∠TPO = 90o (Theorem 32.1)
• Thus ∠TPQ = ∠TPO - ∠OPQ = (90 -θ)
2. TQ is the tangent drawn from an external point T to a point Q on the circle
• OQ is the line joining that Q to the center O
• So ∠TQO = 90o (Theorem 32.1)
• Thus ∠TQP = ∠TQO - ∠OQP = (90 - ∠OQP)
3. OPQ is an isosceles triangle (∵ OP = OQ, the radii of the same circle)
• Base angles of an isosceles triangle are equal. So we get:
• ∠OQP = ∠OPQ = θ.
4. Substituting this value of ∠OQP in (2) we get:
∠TQP = (90-θ)
5. Consider triangle TPQ
• ∠PTQ = [180 -(∠TPQ+∠TQP)]
= [180 - {(90-θ)+(90-θ)}]
= [180 - {2 × (90-θ)}]
= [180 - {180 - 2θ}] = [180 - 180 + 2θ] = 2θ
Solved example 32.4
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T See below
Find the length TP
Solution:
1. Draw OT. Let it intersect PQ at R (fig.32.13.b)
• TP and TQ are tangents from the same point T.
• Then ∠PTO = ∠QTO (Theorem 32.3 part(ii)). These two angles are shown in blue color in fig.32.13.b
2. Also TP = TQ (Theorem 32.2)
• So PTQ is an isosceles triangle
• In this isosceles triangle, we have: ∠PTO = ∠QTO
• That means, TO is the angle bisector of PTQ
• In the isosceles triangle, if TO is the angle bisector of the apex angle, then that TO will be the perpendicular bisector of the base
• So TO is the perpendicular bisector of PQ
• Thus we get: PQ = QR = 4 cm
3. Since TO is perpendicular to PQ, we get four right triangles:
⊿TPR, ⊿OPR, ⊿TQR and ⊿OQR
4. Applying Pythagoras theorem in ⊿OPR, we get:
OR2 = OP2 - PR2 ⟹ OR2 = 52 - 42 ⟹ OR2 = 25 - 16 = 9 ⟹ OR = √9 = 3 cm
5. Applying Pythagoras theorem in ⊿TPR, we get:
TR2 = TP2 - PR2 ⟹ TR2 = TP2 - 42
6. Applying Pythagoras theorem in ⊿OPT, we get:
TP2 = OT2 - OP2 ⟹ TP2 = (TR+OR)2 - 52
⟹ TP2 = (TR2 + 2 × TR × OR + OR2 - 25)
⟹ TP2 = (TR2 + 2 × TR × 3 + 32 -25)
⟹ TP2 = (TR2 + 6TR + 9 -25)
⟹ TP2 = (TR2 + 6TR -16)
7. But from (5), we have: TR2 = TP2 - 42 = TP2 - 16
Substituting this in (6) we get:
TP2 = (TP2 - 16 + 6TR -16) ⟹ 6TR = 32 ⟹ TR = 32⁄6 = 16⁄3
8. Substituting this value of TR in (5) we get:
(16⁄3)2 = TP2 - 42 ⟹ TP2 = (16⁄3)2 + 16
⟹ TP2 = 256⁄9+ 16 ⟹ TP2 = 16(16+9)⁄9 = 16(25)⁄9
TP = √[16(25)⁄9] = 4(5)⁄3 = 20⁄3 cm
Solved example 32.5
Prove that the tangents drawn to a circle at the two ends of a diameter are parallel
Solution:
1. Fig.32.14(a) below shows a rough sketch. The diameter is shown in green colour. Tangents are drawn at it's ends in red colour.
• We have to prove that the red lines are parallel
2. The diameter and tangents are given names in fig.32.14(b)
• PQ is the diameter
• Tangent at P is named as AB
• Tangent at Q is named as CD
3. Extend PQ beyond the tangents
• Now we have two red lines cut by a green transversal
4. If AB is to be a tangent at P, ∠APO = 90o
• If CD is to be a tangent at Q, ∠CQO = 90o
• But ∠APO and ∠CQO are corresponding angles. If corresponding angles are equal, the two red lines will definitely be parallel
Solved example 32.3
Two tangents TP and TQ are drawn to a circle with center O from an external point T (fig.32.12.a).
Prove that ∠PTQ = 2 ∠OPQ
Solution:
• Let ∠OPQ = θ (fig.32.12.b). Then we have to prove that ∠PTQ = 2θ
We will write the steps:
1. TP is the tangent drawn from an external point T to a point P on the circle
• OP is the line joining that P to the center O
• So ∠TPO = 90o (Theorem 32.1)
• Thus ∠TPQ = ∠TPO - ∠OPQ = (90 -θ)
2. TQ is the tangent drawn from an external point T to a point Q on the circle
• OQ is the line joining that Q to the center O
• So ∠TQO = 90o (Theorem 32.1)
• Thus ∠TQP = ∠TQO - ∠OQP = (90 - ∠OQP)
3. OPQ is an isosceles triangle (∵ OP = OQ, the radii of the same circle)
• Base angles of an isosceles triangle are equal. So we get:
• ∠OQP = ∠OPQ = θ.
4. Substituting this value of ∠OQP in (2) we get:
∠TQP = (90-θ)
5. Consider triangle TPQ
• ∠PTQ = [180 -(∠TPQ+∠TQP)]
= [180 - {(90-θ)+(90-θ)}]
= [180 - {2 × (90-θ)}]
= [180 - {180 - 2θ}] = [180 - 180 + 2θ] = 2θ
Solved example 32.4
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T See below
 |
| Fig.32.13 |
Solution:
1. Draw OT. Let it intersect PQ at R (fig.32.13.b)
• TP and TQ are tangents from the same point T.
• Then ∠PTO = ∠QTO (Theorem 32.3 part(ii)). These two angles are shown in blue color in fig.32.13.b
2. Also TP = TQ (Theorem 32.2)
• So PTQ is an isosceles triangle
• In this isosceles triangle, we have: ∠PTO = ∠QTO
• That means, TO is the angle bisector of PTQ
• In the isosceles triangle, if TO is the angle bisector of the apex angle, then that TO will be the perpendicular bisector of the base
• So TO is the perpendicular bisector of PQ
• Thus we get: PQ = QR = 4 cm
3. Since TO is perpendicular to PQ, we get four right triangles:
⊿TPR, ⊿OPR, ⊿TQR and ⊿OQR
4. Applying Pythagoras theorem in ⊿OPR, we get:
OR2 = OP2 - PR2 ⟹ OR2 = 52 - 42 ⟹ OR2 = 25 - 16 = 9 ⟹ OR = √9 = 3 cm
5. Applying Pythagoras theorem in ⊿TPR, we get:
TR2 = TP2 - PR2 ⟹ TR2 = TP2 - 42
6. Applying Pythagoras theorem in ⊿OPT, we get:
TP2 = OT2 - OP2 ⟹ TP2 = (TR+OR)2 - 52
⟹ TP2 = (TR2 + 2 × TR × OR + OR2 - 25)
⟹ TP2 = (TR2 + 2 × TR × 3 + 32 -25)
⟹ TP2 = (TR2 + 6TR + 9 -25)
⟹ TP2 = (TR2 + 6TR -16)
7. But from (5), we have: TR2 = TP2 - 42 = TP2 - 16
Substituting this in (6) we get:
TP2 = (TP2 - 16 + 6TR -16) ⟹ 6TR = 32 ⟹ TR = 32⁄6 = 16⁄3
8. Substituting this value of TR in (5) we get:
(16⁄3)2 = TP2 - 42 ⟹ TP2 = (16⁄3)2 + 16
⟹ TP2 = 256⁄9+ 16 ⟹ TP2 = 16(16+9)⁄9 = 16(25)⁄9
TP = √[16(25)⁄9] = 4(5)⁄3 = 20⁄3 cm
Solved example 32.5
Prove that the tangents drawn to a circle at the two ends of a diameter are parallel
Solution:
1. Fig.32.14(a) below shows a rough sketch. The diameter is shown in green colour. Tangents are drawn at it's ends in red colour.
 |
| Fig.30.14 |
2. The diameter and tangents are given names in fig.32.14(b)
• PQ is the diameter
• Tangent at P is named as AB
• Tangent at Q is named as CD
3. Extend PQ beyond the tangents
• Now we have two red lines cut by a green transversal
4. If AB is to be a tangent at P, ∠APO = 90o
• If CD is to be a tangent at Q, ∠CQO = 90o
• But ∠APO and ∠CQO are corresponding angles. If corresponding angles are equal, the two red lines will definitely be parallel
No comments:
Post a Comment