In the previous section we saw the two tangents from an exterior point. We also saw some solved examples. In this section we will see some more advanced details related to tangents.
We have seen the case when two chords of a circle intersect outside the circle. See fig.27.68 in chapter 27. That fig. is shown again below:
Based on the fig., we derived the following result:
PA × PB = PC × PD
■ But what if one of them is a tangent?
Such a situation is shown in fig.32.46(a) below:
Let us do an analysis. We will write the steps:
1. One line from P cuts the circle at A and B
• The other line from P is a tangent, which touches the circle at C
2. We want the relation between these two line.
• For that, draw the chords AC and BC. This is shown in fig.b
• Now, ∠PCA is the angle between the chord AC and the tangent on the 'tangent side'
• ∠ABC is the angle made by the chord AC (at a point on the circle) on the non-tangent side
• Those two angles will be the same. (Theorem 32.7)
3. Separate the two triangles ΔPAC and ΔPBC as shown in fig.c
• Angle at P is the same in both triangles
• Angle at C in ΔPAC = Angle at B in ΔPBC
■ Two angles are same. So the third angle will also be the same. That is.,
• Angle at A in ΔPAC = Angle at C in ΔPBC
4. So all the three angles are the same.
The two triangles ΔPAC and ΔPBC are similar.
5. Now we apply a special property that is applicable to any two similar triangles (Details here):
= side opposite A in ΔPAC⁄side opposite C in ΔPBC
= side opposite C in ΔPAC⁄side opposite B in ΔPBC
7. So we get: AC⁄BC = PC⁄PB = PA⁄PC
In the next section, we will see more details about Tangents.
We have seen the case when two chords of a circle intersect outside the circle. See fig.27.68 in chapter 27. That fig. is shown again below:
 |
| Fig.27.68 |
PA × PB = PC × PD
■ But what if one of them is a tangent?
Such a situation is shown in fig.32.46(a) below:
 |
| Fig.32.46 |
1. One line from P cuts the circle at A and B
• The other line from P is a tangent, which touches the circle at C
2. We want the relation between these two line.
• For that, draw the chords AC and BC. This is shown in fig.b
• Now, ∠PCA is the angle between the chord AC and the tangent on the 'tangent side'
• ∠ABC is the angle made by the chord AC (at a point on the circle) on the non-tangent side
• Those two angles will be the same. (Theorem 32.7)
3. Separate the two triangles ΔPAC and ΔPBC as shown in fig.c
• Angle at P is the same in both triangles
• Angle at C in ΔPAC = Angle at B in ΔPBC
■ Two angles are same. So the third angle will also be the same. That is.,
• Angle at A in ΔPAC = Angle at C in ΔPBC
4. So all the three angles are the same.
The two triangles ΔPAC and ΔPBC are similar.
5. Now we apply a special property that is applicable to any two similar triangles (Details here):
side opposite smallest angle in ΔPAC⁄side opposite smallest angle in ΔPBC
= side opposite medium angle in ΔPAC⁄side opposite medium angle in ΔPBC
= side opposite largest angle in ΔPAC⁄side opposite largest angle in ΔPBC
5. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔPAC = Smallest angle in ΔPBC
• Medium angle in ΔPAC = Medium angle in ΔPBC
• Largest angle in ΔPAC = Largest angle in ΔPBC
6. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
side opposite P in ΔPAC⁄side opposite P in ΔPBC= side opposite A in ΔPAC⁄side opposite C in ΔPBC
= side opposite C in ΔPAC⁄side opposite B in ΔPBC
7. So we get: AC⁄BC = PC⁄PB = PA⁄PC
8. Take the second and the third ratios. We get:
• PC⁄PB = PA⁄PC . From this we get:
PA × PB = PC2.
This is a very useful result. We can write it as a theorem.
Theorem 32.9:
1. Two lines are drawn from an exterior point P towards a circle
• One of them cuts the circle at two points A and B
• The other is a tangent which touches the circle at C
2. Consider the line which cuts the circle
(i) It's whole length is PB
(ii) It's length outside the circle is PA
3. Take the product. That is:
(Whole length) × (Length outside)
■ This product will be equal to the square of the length of the tangent
Let us see some solved examples:
Solved example 32.21
In the fig.32.47(a) below, AB is a diameter and P is a point on AB extended.
A tangent from P touches the circle at Q. What is the radius of the circle?
Solution:
1. We can apply the above theorem 32.9. In this case we can write:
• Two lines are drawn from an exterior point P towards a circle
• One of them cuts the circle at two points A and B
• The other is a tangent which touches the circle at Q
2. So we can write: PA × PB = PQ2
⟹ 8 × PB = 42 ⟹ PB = 16⁄8 = 2 cm
3. Thus we get: AB = AP-PB = 8-2 = 6 cm
• So radius = 6⁄2 = 3cm
Solved example 32.22
In the fig.32.47(b) above, the line joining two points A and B on a circle is extended by 4 cm to reach P and a tangent is drawn from P. Length of that tangent PQ is 6 cm.
In the fig.32.47(c), P is further extended along the same line AB by 1 cm. Then a tangent PR is drawn. What is the length of the tangent PR?
Solution:
1. Applying theorem 32.9 to fig.b, we get:
PA × PB = PQ2 ⟹ PA × 4 = 62 ⟹ PA = 36⁄4 = 9 cm
• So we get PA in fig.b. We can use it in fig.c.
But since there is a further extension of 1 cm, the new PA in fig.c is (9+1) = 10 cm
2. Applying theorem 32.9 to fig.c, we get:
PA × PB = PR2 ⟹ 10 × 5 = PR2 ⟹ PR = √50 = √[2×25] = √2 × √25 = 5√2 cm
Let us see a practical application of theorem 32.9:
• We derived theorem 32.9 based on fig.32.46(a) above. The same fig. is shown again as fig.32.48(a) below:
1. Consider the segment PC. Let it be the side of a square.
• If we are given any one side of a square, we can easily construct the whole square. The completed square is named as PQRC in fig.b
• Area of PQRC is PC2.
2. Draw a line perpendicular to PB
• With P as center and PB as radius, draw an arc cutting the perpendicular line at S. Then PS = PB
3. Consider PA and PS as the two adjacent sides of a rectangle.
• Knowing the two adjacent sides, we can easily complete the whole rectangle.
• It is named as PSTA in fig.c
• Area of rectangle PSTA = PA × PS = PA × PB (∵ PS = PB)
4. But from theorem 32.9, we have: PA × PB = PC2.
So we can write:
■ Area of rectangle PSTA = Area of square PQRC
• This gives us an effective method to draw a square which has the same area as a given rectangle and vice versa
Let us see some solved example:
Solved example 32.23
Draw a rectangle of one side 6 cm and area equal to that of a square of side 5 cm
Solution:
Step 1 (fig.32.49.a):
1. Draw a horizontal line AB, 6 cm in length.
2. Draw a perpendicular bisector of AB and thus mark the center O1 of AB
3. With O1 as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 5 cm, cutting the semicircle at C.
• Then ∠ACB will be 90o. (Details here)
Step 2 (fig.32.49.b):
1. Draw a perpendicular bisector of BC and thus mark the center O2 of BC
2. With O2 as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.49.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get:
AD × AB = AC2 ⟹ AD × AE = AC2 ⟹ AD × AE = 52
2. So Area of rectangle AEFD is same as that of a square of side 5 cm.
3. Thus AEFD is the required rectangle.
Solved example 32.24
Draw a square of area 25 sq.cm. Draw a rectangle of one side 8 cm and area equal to that of the square.
Solution:
Step 1 (fig.32.50.a):
1. Draw a horizontal line AB, 8 cm in length.
2. Draw a perpendicular bisector of AB and thus mark the center O1 of AB
3. With O1 as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 5 cm, cutting the semicircle at C.
• Then ∠ACB will be 90o. (Details here)
Step 2 (fig.32.50.b):
1. Draw a perpendicular bisector of BC and thus mark the center O2 of BC
2. With O2 as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.50.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get:
AD × AB = AC2 ⟹ AD × AE = AC2 ⟹ AD × AE = 52
2. So Area of rectangle AEFD is same as that of a square of side 5 cm.
3. Thus AEFD is the required rectangle
Solved example 32.25
Draw a rectangle of one side 6 cm and area equal to that of a square of side 4 cm
Solution:
Step 1 (fig.32.51.a):
1. Draw a horizontal line AB, 6 cm in length.
2. Draw a perpendicular bisector of AB and thus mark the center O1 of AB
3. With O1 as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 4 cm, cutting the semicircle at C.
• Then ∠ACB will be 90o. (Details here)
Step 2 (fig.32.51.b):
1. Draw a perpendicular bisector of BC and thus mark the center O2 of BC
2. With O2 as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.51.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get:
AD × AB = AC2 ⟹ AD × AE = AC2 ⟹ AD × AE = 42
2. So Area of rectangle AEFD is same as that of a square of side 4 cm.
3. Thus AEFD is the required rectangle.
Solved example 32.26
Draw a rectangle of one side 7 cm and area 36 sq.cm
Solution:
Step 1 (fig.32.52.a):
1. Draw a horizontal line AB, 7 cm in length.
2. Draw a perpendicular bisector of AB and thus mark the center O1 of AB
3. With O1 as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 6 cm, cutting the semicircle at C.
• Then ∠ACB will be 90o. (Details here)
Step 2 (fig.32.52.b):
1. Draw a perpendicular bisector of BC and thus mark the center O2 of BC
2. With O2 as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.52.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get:
AD × AB = AC2 ⟹ AD × AE = AC2 ⟹ AD × AE = 62
2. So Area of rectangle AEFD is same as that of a square of side 6 cm.
3. Thus AEFD is the required rectangle.
• PC⁄PB = PA⁄PC . From this we get:
PA × PB = PC2.
This is a very useful result. We can write it as a theorem.
Theorem 32.9:
1. Two lines are drawn from an exterior point P towards a circle
• One of them cuts the circle at two points A and B
• The other is a tangent which touches the circle at C
2. Consider the line which cuts the circle
(i) It's whole length is PB
(ii) It's length outside the circle is PA
3. Take the product. That is:
(Whole length) × (Length outside)
■ This product will be equal to the square of the length of the tangent
Let us see some solved examples:
Solved example 32.21
In the fig.32.47(a) below, AB is a diameter and P is a point on AB extended.
 |
| Fig.32.47 |
Solution:
1. We can apply the above theorem 32.9. In this case we can write:
• Two lines are drawn from an exterior point P towards a circle
• One of them cuts the circle at two points A and B
• The other is a tangent which touches the circle at Q
2. So we can write: PA × PB = PQ2
⟹ 8 × PB = 42 ⟹ PB = 16⁄8 = 2 cm
3. Thus we get: AB = AP-PB = 8-2 = 6 cm
• So radius = 6⁄2 = 3cm
Solved example 32.22
In the fig.32.47(b) above, the line joining two points A and B on a circle is extended by 4 cm to reach P and a tangent is drawn from P. Length of that tangent PQ is 6 cm.
In the fig.32.47(c), P is further extended along the same line AB by 1 cm. Then a tangent PR is drawn. What is the length of the tangent PR?
Solution:
1. Applying theorem 32.9 to fig.b, we get:
PA × PB = PQ2 ⟹ PA × 4 = 62 ⟹ PA = 36⁄4 = 9 cm
• So we get PA in fig.b. We can use it in fig.c.
But since there is a further extension of 1 cm, the new PA in fig.c is (9+1) = 10 cm
2. Applying theorem 32.9 to fig.c, we get:
PA × PB = PR2 ⟹ 10 × 5 = PR2 ⟹ PR = √50 = √[2×25] = √2 × √25 = 5√2 cm
Let us see a practical application of theorem 32.9:
• We derived theorem 32.9 based on fig.32.46(a) above. The same fig. is shown again as fig.32.48(a) below:
 |
| `Fig.32.48 |
• If we are given any one side of a square, we can easily construct the whole square. The completed square is named as PQRC in fig.b
• Area of PQRC is PC2.
2. Draw a line perpendicular to PB
• With P as center and PB as radius, draw an arc cutting the perpendicular line at S. Then PS = PB
3. Consider PA and PS as the two adjacent sides of a rectangle.
• Knowing the two adjacent sides, we can easily complete the whole rectangle.
• It is named as PSTA in fig.c
• Area of rectangle PSTA = PA × PS = PA × PB (∵ PS = PB)
4. But from theorem 32.9, we have: PA × PB = PC2.
So we can write:
■ Area of rectangle PSTA = Area of square PQRC
• This gives us an effective method to draw a square which has the same area as a given rectangle and vice versa
Let us see some solved example:
Solved example 32.23
Draw a rectangle of one side 6 cm and area equal to that of a square of side 5 cm
Solution:
Step 1 (fig.32.49.a):
 |
| Fig.32.49 |
2. Draw a perpendicular bisector of AB and thus mark the center O1 of AB
3. With O1 as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 5 cm, cutting the semicircle at C.
• Then ∠ACB will be 90o. (Details here)
Step 2 (fig.32.49.b):
1. Draw a perpendicular bisector of BC and thus mark the center O2 of BC
2. With O2 as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.49.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get:
AD × AB = AC2 ⟹ AD × AE = AC2 ⟹ AD × AE = 52
2. So Area of rectangle AEFD is same as that of a square of side 5 cm.
3. Thus AEFD is the required rectangle.
Solved example 32.24
Draw a square of area 25 sq.cm. Draw a rectangle of one side 8 cm and area equal to that of the square.
Solution:
Step 1 (fig.32.50.a):
 |
| Fig.32.50 |
2. Draw a perpendicular bisector of AB and thus mark the center O1 of AB
3. With O1 as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 5 cm, cutting the semicircle at C.
• Then ∠ACB will be 90o. (Details here)
Step 2 (fig.32.50.b):
1. Draw a perpendicular bisector of BC and thus mark the center O2 of BC
2. With O2 as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.50.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get:
AD × AB = AC2 ⟹ AD × AE = AC2 ⟹ AD × AE = 52
2. So Area of rectangle AEFD is same as that of a square of side 5 cm.
3. Thus AEFD is the required rectangle
Solved example 32.25
Draw a rectangle of one side 6 cm and area equal to that of a square of side 4 cm
Solution:
Step 1 (fig.32.51.a):
 |
| Fig.32.51 |
2. Draw a perpendicular bisector of AB and thus mark the center O1 of AB
3. With O1 as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 4 cm, cutting the semicircle at C.
• Then ∠ACB will be 90o. (Details here)
Step 2 (fig.32.51.b):
1. Draw a perpendicular bisector of BC and thus mark the center O2 of BC
2. With O2 as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.51.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get:
AD × AB = AC2 ⟹ AD × AE = AC2 ⟹ AD × AE = 42
2. So Area of rectangle AEFD is same as that of a square of side 4 cm.
3. Thus AEFD is the required rectangle.
Solved example 32.26
Draw a rectangle of one side 7 cm and area 36 sq.cm
Solution:
Step 1 (fig.32.52.a):
 |
| Fig.32.52 |
2. Draw a perpendicular bisector of AB and thus mark the center O1 of AB
3. With O1 as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 6 cm, cutting the semicircle at C.
• Then ∠ACB will be 90o. (Details here)
Step 2 (fig.32.52.b):
1. Draw a perpendicular bisector of BC and thus mark the center O2 of BC
2. With O2 as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.52.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get:
AD × AB = AC2 ⟹ AD × AE = AC2 ⟹ AD × AE = 62
2. So Area of rectangle AEFD is same as that of a square of side 6 cm.
3. Thus AEFD is the required rectangle.
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