In the previous section we saw some solved examples demonstrating the relations between tangents and chords. In this section we will see some constructions related to tangents.
Let us recall three important points that we saw earlier in this chapter:
Case 1: No tangent can be drawn through a point inside the circle
Case 2: Only one tangent can be drawn through a point on the circle
■ 'Only one' indicates that, it is unique.
• It can be explained as follows:
♦ Take any point on a circle. That particular point will have a 'particular direction' in which a tangent can be drawn
♦ Through that point, a tangent cannot be drawn in any other direction
• This is shown in fig.32.38(a) below:
Case 3: Only two tangents can be drawn through a point outside the circle
■ 'Only two' indicates that those two are unique.
• It can be explained as follows:
♦ Take any point on the exterior of the circle. That particular point will have 'two particular directions' in which tangents can be drawn.
♦ Through that point, tangents cannot be drawn in any other directions
• This is shown in fig.32.38(b) above
• We know how to draw a tangent in case 2. That is., we know the construction method for case 2.
• But we have not yet seen the construction for case 3. Let us now try to do such a construction:
• In fig.32.39(a) below, a circle with centre 'O' and radius 2 cm is shown.
• A point 'T' is marked at a distance of 5 cm form 'O'
• We want to draw the two tangents from T
• Let the two tangent points be P and Q. Let us try the tangent TP first:
• If we know the position of P, we can draw the radial line OP. Then all we have to do is, draw a line through P, perpendicular to OP. So ∠TPO will be 90o. This is shown in fig(b).
• But we do not know where P is. So we cannot draw TP. Same is the case with TQ
• So we will need the position of P and Q first. For that, the following analysis is helpful:
1. In fig(c), a cyan circle is drawn with OT as diameter.
• Since OT is a diameter, we will get two semicircles. One above and the other below OT
2. Consider any point P1 on the bottom semicircle. Draw P1T and P1O
• Since OT is a diameter, ∠TP1O will be 90o. (Details here)
3. Like P1, we can mark any number of points that we like. P2, P3, P4 . . ,
• All those points can be connected to T and O
• Thus we will get a large number of angles: ∠TP1O, ∠TP2O, ∠TP3O . . ,
• All those angles will be equal to 90o
4. But only one of them is of use to us. Which is it?
• In the above angles, P1, P2, P3 etc., are the 'vertices of the angles' (Definition can be seen here)
• For the angle to be of use to us, it's vertex should be on our original yellow circle with center 'O'
• Such a point lies on the intersection of the yellow and cyan circles. It is marked as P
• So we have ∠TPO = 90o. Thus TP is the tangent drawn from T
5. In a similar way, ∠TQO is also equal to 90o. Thus TQ is the other tangent
• So we have a method to draw the two tangents from an exterior point T. Let us write the steps:
Step 1: Draw OT (cyan line in fig.32.40.a below)
• Draw the perpendicular bisector (dashed cyan line in fig.a) of OT
♦ This is to find the midpoint O1 of OT
Step 2: With O1 as center and O1T as radius, draw the cyan circle in fig.b. Then OT will be it's diameter.
• Mart the points of intersection of the two circles as P and Q
Step 3: Draw TP and TQ. They are the required tangents. This is shown in fig(c)
• In the fig.32.39(c) above, we have a right triangle: ⊿OPT
♦ OT is the hypotenuse since it is opposite the 90o angle
• Applying Pythagoras theorem, we get:
TP2 = OT2 - OP2 ⟹ TP = √[OT2 - OP2]
• This is same as TQ
• In our present problem, we have: OT = 5 cm and OP = 2 cm
• So we get: TP = TQ = √[OT2 - OP2] = √[52 - 22] = √[25 - 4] = √[21] cm
Another interesting result:
• In fig.32.41(a) below, a yellow circle is drawn with center at 'O'.
• Four points P, Q R and S are marked on the circle and red tangents are drawn through them.
• The tangents intersect at four points A, B, C and D. So ABCD is a quadrilateral
• Based on the above information, let us write some steps:
1. Tangents through P and S meet at A. So AS will be equal to AP
Let us put AS = PS = a. This is shown in fig.b
• Tangents through P and Q meet at B. So BP will be equal to BQ
Let us put BP = BQ = b
• Tangents through Q and R meet at C. So CQ will be equal to CR
Let us put CQ = CR = c
• Tangents through R and S meet at D. So DR will be equal to DS
Let us put DR = DS = d
2. So length of side AB = (a+b)
• Length of side BC = (b+c)
• Length of side CD = (c+d)
• Length of side AD = (d+a)
3. Let us add opposite sides:
• AB + CD = [(a+b) + (c + d)] = [a+b+c+d]
• BC + AD = [(b+c) + (a + d)] = [a+b+c+d]
■ Both are same
We can write the above result in the form of a theorem:
Theorem 32.8
• Four points are marked on the circle and tangents are drawn through them
• The four tangents meet at four points to give the four vertices of a quadrilateral
• Sum of the opposite sides of that quadrilateral will be equal
Now we will see a solved example:
Solved example 32.16
Fig.32.42(a) below shows a triangle formed by three tangents to a circle.
Calculate the length of each tangent from the corner of the triangle to point of contact
Solution:
• Let the triangle be ABC and the tangent points be P, Q and R. This is shown in fig(b)
1. Tangents through P and R meet at A. So AS will be equal to AP
Let us put AP = AR = a.
• Tangents through P and Q meet at B. So BP will be equal to BQ
Let us put BP = BQ = b
• Tangents through Q and R meet at C. So CQ will be equal to CR
Let us put CQ = CR = c
2. So we can write:
(i) a+b = 7
(ii) b+c = 5
(iii) a+c = 4
• Thus we have 3 unknowns: a, b and c
• We also have three equations
3. From 2(i) we get:
(i) a = 7-b
• Substituting this value of a in 2(iii) we get:
(ii) 7-b+c = 4 ⟹ -b+c = -3 ⟹ c = b-3
• Substituting this value of c in 2(ii) we get:
b+b-3 = 5 ⟹ 2b = 8 ⟹ b = 4 cm
• Substituting this value of b in 2(i) we get:
a = 7-4 = 3 cm
• Substituting this value of a in 2(iii) we get:
3+c = 4 ⟹ c = 1 cm
4. Thus we get the required answers:
(i) Length of the tangents AB and AC from the corner A to the tangent points P and R
= AP = AR = a = 3 cm
(ii) Length of the tangents AB and BC from the corner B to the tangent points P and Q
= BP = BQ = b = 4 cm
(iii) Length of the tangents CA and CB from the corner C to the tangent points R and Q
= CR = CQ = c = 1 cm
In the next section, we will see a few more solved examples.
Let us recall three important points that we saw earlier in this chapter:
Case 1: No tangent can be drawn through a point inside the circle
Case 2: Only one tangent can be drawn through a point on the circle
■ 'Only one' indicates that, it is unique.
• It can be explained as follows:
♦ Take any point on a circle. That particular point will have a 'particular direction' in which a tangent can be drawn
♦ Through that point, a tangent cannot be drawn in any other direction
• This is shown in fig.32.38(a) below:
 |
| Fig.32.38 |
■ 'Only two' indicates that those two are unique.
• It can be explained as follows:
♦ Take any point on the exterior of the circle. That particular point will have 'two particular directions' in which tangents can be drawn.
♦ Through that point, tangents cannot be drawn in any other directions
• This is shown in fig.32.38(b) above
• We know how to draw a tangent in case 2. That is., we know the construction method for case 2.
• But we have not yet seen the construction for case 3. Let us now try to do such a construction:
• In fig.32.39(a) below, a circle with centre 'O' and radius 2 cm is shown.
 |
| Fig.32.39 |
• We want to draw the two tangents from T
• Let the two tangent points be P and Q. Let us try the tangent TP first:
• If we know the position of P, we can draw the radial line OP. Then all we have to do is, draw a line through P, perpendicular to OP. So ∠TPO will be 90o. This is shown in fig(b).
• But we do not know where P is. So we cannot draw TP. Same is the case with TQ
• So we will need the position of P and Q first. For that, the following analysis is helpful:
1. In fig(c), a cyan circle is drawn with OT as diameter.
• Since OT is a diameter, we will get two semicircles. One above and the other below OT
2. Consider any point P1 on the bottom semicircle. Draw P1T and P1O
• Since OT is a diameter, ∠TP1O will be 90o. (Details here)
3. Like P1, we can mark any number of points that we like. P2, P3, P4 . . ,
• All those points can be connected to T and O
• Thus we will get a large number of angles: ∠TP1O, ∠TP2O, ∠TP3O . . ,
• All those angles will be equal to 90o
4. But only one of them is of use to us. Which is it?
• In the above angles, P1, P2, P3 etc., are the 'vertices of the angles' (Definition can be seen here)
• For the angle to be of use to us, it's vertex should be on our original yellow circle with center 'O'
• Such a point lies on the intersection of the yellow and cyan circles. It is marked as P
• So we have ∠TPO = 90o. Thus TP is the tangent drawn from T
5. In a similar way, ∠TQO is also equal to 90o. Thus TQ is the other tangent
• So we have a method to draw the two tangents from an exterior point T. Let us write the steps:
Step 1: Draw OT (cyan line in fig.32.40.a below)
 |
| Fig.32.40 |
♦ This is to find the midpoint O1 of OT
Step 2: With O1 as center and O1T as radius, draw the cyan circle in fig.b. Then OT will be it's diameter.
• Mart the points of intersection of the two circles as P and Q
Step 3: Draw TP and TQ. They are the required tangents. This is shown in fig(c)
Length of tangents
• The above discussion gives us an easy method to calculate the length of the tangents drawn from an exterior point• In the fig.32.39(c) above, we have a right triangle: ⊿OPT
♦ OT is the hypotenuse since it is opposite the 90o angle
• Applying Pythagoras theorem, we get:
TP2 = OT2 - OP2 ⟹ TP = √[OT2 - OP2]
• This is same as TQ
• In our present problem, we have: OT = 5 cm and OP = 2 cm
• So we get: TP = TQ = √[OT2 - OP2] = √[52 - 22] = √[25 - 4] = √[21] cm
Another interesting result:
• In fig.32.41(a) below, a yellow circle is drawn with center at 'O'.
 |
| Fig.32.41 |
• The tangents intersect at four points A, B, C and D. So ABCD is a quadrilateral
• Based on the above information, let us write some steps:
1. Tangents through P and S meet at A. So AS will be equal to AP
Let us put AS = PS = a. This is shown in fig.b
• Tangents through P and Q meet at B. So BP will be equal to BQ
Let us put BP = BQ = b
• Tangents through Q and R meet at C. So CQ will be equal to CR
Let us put CQ = CR = c
• Tangents through R and S meet at D. So DR will be equal to DS
Let us put DR = DS = d
2. So length of side AB = (a+b)
• Length of side BC = (b+c)
• Length of side CD = (c+d)
• Length of side AD = (d+a)
3. Let us add opposite sides:
• AB + CD = [(a+b) + (c + d)] = [a+b+c+d]
• BC + AD = [(b+c) + (a + d)] = [a+b+c+d]
■ Both are same
We can write the above result in the form of a theorem:
Theorem 32.8
• Four points are marked on the circle and tangents are drawn through them
• The four tangents meet at four points to give the four vertices of a quadrilateral
• Sum of the opposite sides of that quadrilateral will be equal
Now we will see a solved example:
Solved example 32.16
Fig.32.42(a) below shows a triangle formed by three tangents to a circle.
 |
| Fig.32.42 |
Solution:
• Let the triangle be ABC and the tangent points be P, Q and R. This is shown in fig(b)
1. Tangents through P and R meet at A. So AS will be equal to AP
Let us put AP = AR = a.
• Tangents through P and Q meet at B. So BP will be equal to BQ
Let us put BP = BQ = b
• Tangents through Q and R meet at C. So CQ will be equal to CR
Let us put CQ = CR = c
2. So we can write:
(i) a+b = 7
(ii) b+c = 5
(iii) a+c = 4
• Thus we have 3 unknowns: a, b and c
• We also have three equations
3. From 2(i) we get:
(i) a = 7-b
• Substituting this value of a in 2(iii) we get:
(ii) 7-b+c = 4 ⟹ -b+c = -3 ⟹ c = b-3
• Substituting this value of c in 2(ii) we get:
b+b-3 = 5 ⟹ 2b = 8 ⟹ b = 4 cm
• Substituting this value of b in 2(i) we get:
a = 7-4 = 3 cm
• Substituting this value of a in 2(iii) we get:
3+c = 4 ⟹ c = 1 cm
4. Thus we get the required answers:
(i) Length of the tangents AB and AC from the corner A to the tangent points P and R
= AP = AR = a = 3 cm
(ii) Length of the tangents AB and BC from the corner B to the tangent points P and Q
= BP = BQ = b = 4 cm
(iii) Length of the tangents CA and CB from the corner C to the tangent points R and Q
= CR = CQ = c = 1 cm
No comments:
Post a Comment