In the previous section we saw tangents giving cyclic quadrilaterals. In this section we will learn the relations between tangents and chords.
1. A circle is drawn with center at O. See fig.32.28(a)
2. Two green radial lines OP and OQ are drawn in such a way that, ∠POQ = 100
3. A red tangent is drawn at P. Another red tangent is drawn at Q
4. The red tangents meet at T. So the tangents are named as AT and BT
5. A magenta line is drawn joining P and Q. So the magenta line is a chord
• Clearly, the central angle of the chord is 100o
6. We know that ∠PTQ will be equal to 80o [Theorem 32.5]
• This is shown in fig(b)
7. We also know that PT = QT. [Theorem 32.2]
8. So ΔPTQ is an isosceles triangle. It's base angles are equal. Let them be x
9. So in ΔPTQ, we get: (2x + 80) = 180o ⟹ x = 50o
10. But this 50o is half of the central angle (made by the chord PQ) 100o
Let us see if this is true for any chord:
1. A circle is drawn with center at O. See fig.32.29(a)
2. Two green radial lines OP and OQ are drawn in such a way that, ∠POQ = co
3. A red tangent is drawn at P. Another red tangent is drawn at Q
4. The red tangents meet at T. So the tangents are named as AT and BT
5. A magenta line is drawn joining P and Q. So the magenta line is a chord
• Clearly, the central angle of the chord is co
6. We know that ∠PTQ will be equal to (180-c)o [Theorem 32.5]
7. We also know that PT = QT. [Theorem 32.2]
8. So ΔPTQ is an isosceles triangle. It's base angles are equal. Let them be xo
9. So in ΔPTQ, we get: [2x + (180-c)] = 180o ⟹ (2x - c) = 0 ⟹ x = (c⁄2)o
We can write the above result as a theorem:
Theorem 32.6
• P and Q are two points on the circle
• A tangent is drawn at P
♦ The angle between this tangent and the chord PQ is half of the central angle of the chord
• A tangent is drawn at Q
♦ The angle between this tangent and the chord PQ is also half of the central angle of the chord
Now we will see a more interesting case:
1. The chord PQ in fig.32.29 above, divides the circle into two arcs:
• A minor arc PQ and a major arc PQ
2. Consider any point R on the major arc PQ. This is shown in fig.32.30(a)
• The chord PQ will subtend ∠PRQ on that major arc.
3. We know that ∠PRQ will be equal to half the central angle of the chord PQ. [Theorem 27.4]
• So we get: ∠PRQ = (c⁄2)o
4. But based on the theorem 32.6 that we wrote above, the angle between the chord and the tangent is also (c⁄2)o
5. Thus we get an interesting result: ∠PRQ = ∠TPQ = ∠TQP
Let us see some examples:
1. In fig.32.30(b), PQ is a chord.
• Tangent is drawn at P
• Tangent is drawn at Q
• The two tangents meet at T
2. The meeting point T is on the right side of the chord
• We will call the right side as 'Tangent side of the chord'
• So the left side of the chord can be called: 'Non-tangent side of the chord'
3. A point R is marked on the circle on the 'Non-tangent side of the chord'
■ Wherever we mark R on the major arc, the ∠PRQ will be the same. [Details here]
• In our present case, it is 65o.
4. Then, from what we have seen based on fig.32.30(a), we can directly write:
■ The angle between the chord and tangent will be 65o at both ends of the chord
• That is., ∠TPQ and ∠TQP will be equal to 65
• We do not need to know the position of the center 'O' for writing the above result
5. Note that, ∠TPQ and ∠TQP are the angles on the Tangent side of the chord PQ
• Which are the angles on the Non-tangent side?
They are: ∠APQ and ∠BQP
6. What are their values?
• We know that ∠APQ and ∠TPQ form a linear pair
• So ∠APQ = (180 - ∠TPQ) = (180 - 65) = 115o. This is shown in fig (c)
• Similarly, ∠BQP = 1(80 - ∠TQP) = (180 - 65) = 115o
7. Now consider any point S on the circle on the tangent side of the chord
• PRQS is a cyclic quadrilateral. So ∠PSQ = (180 - ∠PRQ) = (180 - 65) = 115o. This is shown in fig(c)
• Note that, in figs.(b) and (c), we do not know where the center of the circle is. Even then we are able to find various angles
We will write the above findings as a theorem:
Theorem 32.7
1. A chord PQ subtends ∠PSQ at a point S on the circle, on the Tangent side
• At the end P, the chord makes ∠APQ (with the tangent AT) on the Non-tangent side
• At the end Q, the chord makes ∠BQP (with the tangent BT) on the Non-tangent side
■ All the above three angles are equal
2. The chord PQ subtends ∠PRQ at a point R on the Non-tangent side
• At the end P, the chord makes ∠QPT (with the tangent AT) on the Tangent side
• At the end Q, the chord makes ∠TQP (with the tangent BT) on the Tangent side
■ All the above three angles are equal
• In (1), point S is on the Tangent side. The chord angles are on the Non-tangent side
• In (2), point R is on the Non-tangent side. The chord angles are on the Tangent side
Now we will see the practical application of the above theorem 32.7
We know how to draw the tangent at a given point on a circle.
1. Draw a radial line through the given point
2. Draw a line perpendicular to the radial line through the given point. This line is the required tangent.
• So the procedure is simple. But it will not be so simple if the centre of the circle is not known. Because, we will not be able to draw the radial line.
• In such a situation, we can use the above theorem 32.7. Let us see the method:
• In fig.32.31(a) below, a circle is shown and a point P is marked on it. Center of the circle is not given. We are required to draw the tangent at P.
We can use the following procedure:
1. From the point P, draw a convenient chord PQ. This is shown in fig(b)
2. Mark a convenient point R on the circle on the Non-tangent side. Complete the triangle PQR
3. Measure ∠PRQ. Let it be xo
4. Draw a red line through P in such a way that the angle (at P) between the red line and the chord is xo. This is shown in fig(c)
5. Then the red line is the required tangent
An easier method:
• In any case, we will need to draw PQR. This is to obtain the value of x
1. We will draw PQR in such a way that, it is isosceles. For that, with P as centre, draw two arcs cutting the circle at Q and R. This is shown in fig.32.32(a) below:
2. Now complete ΔPQR. It is an isosceles triangle because PR = PQ.
• The base angles at R and Q will be the same.
3. Draw a red line through P in such a way that it is parallel to RQ. Then we have:
• RQ and the red line are two parallel lines. They are cut by a transversal PQ
• So the following two angles are co-interior angles and hence will be equal:
♦ ∠RQP
♦ The angle (at P) between the red line and PQ
4. So we get:
• ∠PRQ = x = ∠RQP = The angle (at P) between the red line and PQ
• So the angle at P is also x. This is shown in fig c.
■ Thus the red line is the tangent at P
An actual construction can be seen in the form of a video presentation here.
In the next section, we will see some solved examples.
1. A circle is drawn with center at O. See fig.32.28(a)
 |
| Fig.32.28 |
3. A red tangent is drawn at P. Another red tangent is drawn at Q
4. The red tangents meet at T. So the tangents are named as AT and BT
5. A magenta line is drawn joining P and Q. So the magenta line is a chord
• Clearly, the central angle of the chord is 100o
6. We know that ∠PTQ will be equal to 80o [Theorem 32.5]
• This is shown in fig(b)
7. We also know that PT = QT. [Theorem 32.2]
8. So ΔPTQ is an isosceles triangle. It's base angles are equal. Let them be x
9. So in ΔPTQ, we get: (2x + 80) = 180o ⟹ x = 50o
10. But this 50o is half of the central angle (made by the chord PQ) 100o
Let us see if this is true for any chord:
1. A circle is drawn with center at O. See fig.32.29(a)
 |
| Fig.32.29 |
3. A red tangent is drawn at P. Another red tangent is drawn at Q
4. The red tangents meet at T. So the tangents are named as AT and BT
5. A magenta line is drawn joining P and Q. So the magenta line is a chord
• Clearly, the central angle of the chord is co
6. We know that ∠PTQ will be equal to (180-c)o [Theorem 32.5]
7. We also know that PT = QT. [Theorem 32.2]
8. So ΔPTQ is an isosceles triangle. It's base angles are equal. Let them be xo
9. So in ΔPTQ, we get: [2x + (180-c)] = 180o ⟹ (2x - c) = 0 ⟹ x = (c⁄2)o
We can write the above result as a theorem:
Theorem 32.6
• P and Q are two points on the circle
• A tangent is drawn at P
♦ The angle between this tangent and the chord PQ is half of the central angle of the chord
• A tangent is drawn at Q
♦ The angle between this tangent and the chord PQ is also half of the central angle of the chord
Now we will see a more interesting case:
1. The chord PQ in fig.32.29 above, divides the circle into two arcs:
• A minor arc PQ and a major arc PQ
2. Consider any point R on the major arc PQ. This is shown in fig.32.30(a)
 |
| Fig.32.30 |
3. We know that ∠PRQ will be equal to half the central angle of the chord PQ. [Theorem 27.4]
• So we get: ∠PRQ = (c⁄2)o
4. But based on the theorem 32.6 that we wrote above, the angle between the chord and the tangent is also (c⁄2)o
5. Thus we get an interesting result: ∠PRQ = ∠TPQ = ∠TQP
Let us see some examples:
1. In fig.32.30(b), PQ is a chord.
• Tangent is drawn at P
• Tangent is drawn at Q
• The two tangents meet at T
2. The meeting point T is on the right side of the chord
• We will call the right side as 'Tangent side of the chord'
• So the left side of the chord can be called: 'Non-tangent side of the chord'
3. A point R is marked on the circle on the 'Non-tangent side of the chord'
■ Wherever we mark R on the major arc, the ∠PRQ will be the same. [Details here]
• In our present case, it is 65o.
4. Then, from what we have seen based on fig.32.30(a), we can directly write:
■ The angle between the chord and tangent will be 65o at both ends of the chord
• That is., ∠TPQ and ∠TQP will be equal to 65
• We do not need to know the position of the center 'O' for writing the above result
5. Note that, ∠TPQ and ∠TQP are the angles on the Tangent side of the chord PQ
• Which are the angles on the Non-tangent side?
They are: ∠APQ and ∠BQP
6. What are their values?
• We know that ∠APQ and ∠TPQ form a linear pair
• So ∠APQ = (180 - ∠TPQ) = (180 - 65) = 115o. This is shown in fig (c)
• Similarly, ∠BQP = 1(80 - ∠TQP) = (180 - 65) = 115o
7. Now consider any point S on the circle on the tangent side of the chord
• PRQS is a cyclic quadrilateral. So ∠PSQ = (180 - ∠PRQ) = (180 - 65) = 115o. This is shown in fig(c)
• Note that, in figs.(b) and (c), we do not know where the center of the circle is. Even then we are able to find various angles
We will write the above findings as a theorem:
Theorem 32.7
1. A chord PQ subtends ∠PSQ at a point S on the circle, on the Tangent side
• At the end P, the chord makes ∠APQ (with the tangent AT) on the Non-tangent side
• At the end Q, the chord makes ∠BQP (with the tangent BT) on the Non-tangent side
■ All the above three angles are equal
2. The chord PQ subtends ∠PRQ at a point R on the Non-tangent side
• At the end P, the chord makes ∠QPT (with the tangent AT) on the Tangent side
• At the end Q, the chord makes ∠TQP (with the tangent BT) on the Tangent side
■ All the above three angles are equal
• In (1), point S is on the Tangent side. The chord angles are on the Non-tangent side
• In (2), point R is on the Non-tangent side. The chord angles are on the Tangent side
Now we will see the practical application of the above theorem 32.7
We know how to draw the tangent at a given point on a circle.
1. Draw a radial line through the given point
2. Draw a line perpendicular to the radial line through the given point. This line is the required tangent.
• So the procedure is simple. But it will not be so simple if the centre of the circle is not known. Because, we will not be able to draw the radial line.
• In such a situation, we can use the above theorem 32.7. Let us see the method:
• In fig.32.31(a) below, a circle is shown and a point P is marked on it. Center of the circle is not given. We are required to draw the tangent at P.
 |
| Fig.32.31 |
1. From the point P, draw a convenient chord PQ. This is shown in fig(b)
2. Mark a convenient point R on the circle on the Non-tangent side. Complete the triangle PQR
3. Measure ∠PRQ. Let it be xo
4. Draw a red line through P in such a way that the angle (at P) between the red line and the chord is xo. This is shown in fig(c)
5. Then the red line is the required tangent
An easier method:
• In any case, we will need to draw PQR. This is to obtain the value of x
1. We will draw PQR in such a way that, it is isosceles. For that, with P as centre, draw two arcs cutting the circle at Q and R. This is shown in fig.32.32(a) below:
 |
| Fig.30.32 |
• The base angles at R and Q will be the same.
3. Draw a red line through P in such a way that it is parallel to RQ. Then we have:
• RQ and the red line are two parallel lines. They are cut by a transversal PQ
• So the following two angles are co-interior angles and hence will be equal:
♦ ∠RQP
♦ The angle (at P) between the red line and PQ
4. So we get:
• ∠PRQ = x = ∠RQP = The angle (at P) between the red line and PQ
• So the angle at P is also x. This is shown in fig c.
■ Thus the red line is the tangent at P
An actual construction can be seen in the form of a video presentation here.
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