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:
 |
| Fig.32.38 |
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.
 |
| Fig.32.39 |
• 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)
 |
| Fig.32.40 |
• 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)
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 |
• 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.
 |
| Fig.32.42 |
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.
In the previous section we saw the relations between tangents and chords. In this section we will see some solved examples based on that discussion.
Solved example 32.11
In fig.32.33 (a) below, the circumcircle of ΔPQR is shown in yellow colour. ∠PQR = 80o and ∠PRQ = 60o.
 |
| Fig.30.33 |
Red tangents are drawn at P, Q and R. The tangents intersect at 3 points to form ΔABC. Find the angles of ΔABC.
Solution:
• Given that ∠PQR = 80oand ∠PRQ = 60o.
♦ So ∠RPQ = [180 -(80+60)] = 40o
1. Consider the side PR. It is a chord of the yellow circle.
• This chord subtends an angle of 80o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 80o
• That is., ∠BPR = ∠BRP = 80o.
• So, in ΔPRB, ∠B = [180 -(80+80)] = 20o. This is shown in fig.(b)
2. Consider the side QR. It is a chord of the yellow circle.
• This chord subtends an angle of 40o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 40o
• That is., ∠QRA = ∠AQR = 40o.
• So, in ΔAQR, ∠A = [180 -(40+40)] = 100o.
3. Consider the side PQ. It is a chord of the yellow circle.
• This chord subtends an angle of 60o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 60o
• That is., ∠CPQ = ∠CQP = 60o.
• So, in ΔCQP, ∠C = [180 -(60+60)] = 60o.
Solved example 32.12
In fig.32.34 (a) below, the circumcircle of ΔPQR is shown in yellow colour.
 |
| Fig.32.34 |
Red tangents are drawn at P, Q and R. The tangents intersect at 3 points to form ΔABC.
∠ABC = 40o and ∠ACB = 60o. Find the angles of ΔPQR.
Solution:
• Given that ∠ABC = 40o and ∠ACB = 60o.
♦ So ∠BAC = [180 -(40+60)] = 80o
1. Consider ΔPRB. It is an isosceles triangle (∵ PB = RB. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠BPR) + 40 = 180 ⟹ ∠BPR = ∠PRB = 70o. This is shown in fig(b)
• Consider the side PR. It is a chord of the yellow circle.
• The angles (at the two ends of that chord) on the tangent side is 70o
• So this chord will subtend an angle of 70o on the non-tangent side
• That is., ∠PQR = 70o.
2. Consider ΔQRA. It is an isosceles triangle (∵ QA = RA. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠AQR) + 80 = 180 ⟹ ∠AQR = ∠ARQ = 50o. This is shown in fig(b)
• Consider the side QR. It is a chord of the yellow circle.
• The angles (at the two ends of that chord) on the tangent side is 50o
• So this chord will subtend an angle of 50o on the non-tangent side
• That is., ∠QPR = 50o.
3. Consider ΔQPC. It is an isosceles triangle (∵ QC = PC. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠CQP) + 60 = 180 ⟹ ∠CQP = ∠QPC = 60o. This is shown in fig(b)
• Consider the side QP. It is a chord of the yellow circle.
• The angles (at the two ends of that chord) on the tangent side is 60o
• So this chord will subtend an angle of 60o on the non-tangent side
• That is., ∠QRP = 60o.
Solved example 32.13
In the fig.32.35(a) given below, PQ, RS and TU are tangents to the circumcircle of ABC. Sort out the equal angles in the fig.
 |
| Fig.32.35 |
Solution:
1. Consider the side AC. It is a chord of the yellow circle.
• This chord subtends ∠ABC on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠ABC
• That is., ∠PAC = ∠TCA = ∠ABC. These are shown in green color in fig(b).
2. Consider the side BC. It is a chord of the yellow circle.
• This chord subtends ∠BAC on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠BAC
• That is., ∠UCB = ∠RBC = ∠BAC. These are shown in magenta color.
3. Consider the side AB. It is a chord of the yellow circle.
• This chord subtends ∠ACB on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠ACB
• That is., ∠QAB = ∠SBA = ∠ACB. These are shown in cyan color.
Solved example 32.14
In the fig.32.36(a) below, O is the center of the circle and AB is the tangent to the circle through Q. Find ∠PQA
 |
| Fig.32.36 |
Solution:
1. Given that, the central angle of chord PQ is 100o. So ∠PRQ = 50o. [Theorem 27.4]
• This is shown in fig(b).
2. In Δ
PQR, consider the side PQ. It is a chord of the yellow circle.
• This chord subtends an angle of 50o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will also be 50o
• That is., ∠PQA = ∠PRQ = 50o.
Solved example 32.15
In the fig.32.37(a) below, O is the center of the circle and QR is a diameter of the circle through Q. AB is a tangent to the circle at P. If ∠BPQ = 50o, find ∠PQR
 |
| Fig.32.37 |
Solution:
1. In Δ
PQR, consider the side PQ. It is a chord of the yellow circle
• The tangent AB is at the end P of the chord
• Draw a tangent CD at the end Q also. This is shown in fig(b)
2. The angle at the end P on the tangent side is given to be 50o
• So the angle at end Q on the tangent side will also be 50o
• That is., ∠DQP = 50o
3. But ∠DQO = 90o (∵ the tangent CD is at the end Q of the diameter QR)
4. So we get: ∠PQR = (∠DQO - ∠DQP) = (90 - 50) = 40o
In the next section, we will see more details about Tangents.
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)
 |
| Fig.32.28 |
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 Δ
PTQ is an isosceles triangle. It's base angles are equal. Let them be x
9. So in Δ
PTQ, 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 |
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 Δ
PTQ is an isosceles triangle. It's base angles are equal. Let them be xo
9. So in Δ
PTQ, 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 |
• 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.
 |
| Fig.32.31 |
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:
 |
| Fig.30.32 |
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.
In the previous section we saw some tangents from an exterior point. In this section we will learn more details.
1. Consider the two tangents AP and BP in fig.32.15 below.
• OA and OB are radii.
♦ The tangents are drawn at the points of intersection of the radii with the circle
 |
| Fig.32.15 |
2. We know that ∠OAP = ∠OBP = 90o
3. Now consider the quadrilateral OAPB. It is formed by the radii and the tangents
4. A and B are two opposite corners. The sum of the angles at those opposite corners = 90+90 = 180o
■ If the sum of opposite angles in any quadrilateral is 180o, that quadrilateral will be a cyclic quadrilateral,
• That is., all the four vertices of that quadrilateral will lie on a circle (We saw details here)
5. So we can draw a circle through all the four vertices of the quadrilateral OAPB. This circle is shown in green colour in fig.32.15(b) above
We can write the above result in the form of a theorem:
Theorem 32.4
• If a quadrilateral has the following four vertices:
♦ First vertex at the centre of a circle
♦ Second vertex at the exterior point from which two tangents are drawn to the circle
♦ Third vertex at one tangent point
♦ Fourth vertex at the other tangent point
■ Then that quadrilateral is cyclic
From the above, we get another useful result:
• In the fig.32.15(b) above, we considered the opposite vertices A and B. What about the other two opposite vertices O and P?
• Obviously, the sum of those two angles must also be 180o (∵ the sum of interior angles of any quadrilaterals is 360o)
This result is also useful for solving problems. We will write it as a theorem:
Theorem 32.5
1. In a circle, two radii are drawn
2. Tangents are drawn at the end point of each of those radii
• These tangents meet at P
3. Then sum of the following two angles is 180o
• Angle at O between the two radii
• Angle at P between the two tangents
Now we will see an application of this theorem in the form of a solved example
Solved example 32.6
Draw an equilateral triangle in such a way that it's all three sides are tangential to a circle of radius 3 cm
Solution:
A rough sketch is shown in fig.32.16(a) below:
 |
| Fig.32.16 |
• The three sides of the triangle are: AB, BC and AC. They are all tangents to the circle of radius 3 cm
• The triangle is to be equilateral. So all it's angles are 60
• Based on this rough fig., we must obtain details to make the actual construction
We will write the steps:
1. From the center O, draw radial lines OP, OQ and OR. This is shown in fig.b
2. Based on theorem 32.5, we get:
• ∠A + ∠ROP = 180 ⟹ 60 + ∠ROP = 180 ⟹ ∠ROP = 120
• ∠B + ∠POQ = 180 ⟹ 60 + ∠POQ = 180 ⟹ ∠POQ = 120
• ∠C + ∠ROQ = 180 ⟹ 60 + ∠ROQ = 180 ⟹ ∠ROQ = 120
• So the three central angles are all 120o. This is marked in fig.c
• We can now begin the actual construction
3. Step 1 is to draw a circle of radius 3 cm. Mark the center as 'O'. Then draw a radial line OP in any convenient direction. This is shown in fig.32.17(a) below:
 |
| Fig.32.17 |
4. Step 2 is to draw the other two radial lines:
• Draw OQ in such a way that it makes an angle of 120o with OP
• Draw OR in such a way that it makes an angle of 120o with OQ
• Then the third angle ∠POR will be naturally 120
This is shown in fig.32.17(b)
5. Step 3 is the final step. In this step we draw the tangents:
• Draw a line perpendicular to OP at P. This is the tangent at P
• Draw a line perpendicular to OQ at Q. This is the tangent at Q
• Draw a line perpendicular to OR at R. This is the tangent at R
■ The three tangents will meet at A, B and C, giving the required equilateral triangle. This is shown in fig.c
Solved example 32.7
Draw a triangle of angles 40o, 60o and 80o in such a way that it's all three sides are tangential to a circle of radius 2.5 cm
Solution:
A rough sketch is shown in fig.32.18(a) below:
 |
| Fig.32.18 |
• The three sides of the triangle are: AB, BC and AC. They are all tangents to the circle of radius 2.5 cm
• The required angles are also marked in the fig.a
• Based on this rough fig., we must obtain details to make the actual construction
We will write the steps:
1. From the center O, draw radial lines OP, OQ and OR. This is shown in fig.b
2. Based on theorem 32.5, we get:
• ∠A + ∠ROP = 180 ⟹ 40 + ∠ROP = 180 ⟹ ∠ROP = 140
• ∠B + ∠POQ = 180 ⟹ 60 + ∠POQ = 180 ⟹ ∠POQ = 120
• ∠C + ∠ROQ = 180 ⟹ 80 + ∠ROQ = 180 ⟹ ∠ROQ = 100
• So the three central angles are 140o, 120o and 100o. This is marked in fig.c
• We can now begin the actual construction
3. Step 1 is to draw a circle of radius 2.5 cm. Mark the center as 'O'. Then draw a radial line OP in any convenient direction. This is shown in fig.32.19(a) below:
 |
| Fig.32.19 |
4. Step 2 is to draw the other two radial lines:
• Draw OQ in such a way that it makes an angle of 120o with OP
• Draw OR in such a way that it makes an angle of 100o with OQ
• Then the third angle ∠POR will be naturally 140o
This is shown in fig.32.19(b)
5. Step 3 is the final step. In this step we draw the tangents:
• Draw a line perpendicular to OP at P. This is the tangent at P
• Draw a line perpendicular to OQ at Q. This is the tangent at Q
• Draw a line perpendicular to OR at R. This is the tangent at R
■ The three tangents will meet at A, B and C, giving the required triangle. This is shown in fig.c
Solved example 32.8
In the fig.32.20(a), the blue triangle is equilateral. It's circumcircle is shown in green colour. All the sides of the red triangle are tangents to the green circle. The tangent points are the vertices of the blue triangle.
 |
| Fig.32.20 |
(i) Prove that the red triangle is also equilateral and it's sides are double that of the blue triangle
(ii) Draw the fig. with sides of the small triangle 3 cm
Solution:
Part 1:
Let ABC be the red triangle and PQR be the blue triangle. This is marked in fig.b
1. Mark the center O of the green circle. Draw radial lines OP, OQ and OR. They are shown in yellow colour
2. Given that the red lines are tangents at the vertices. So the yellow lines are perpendicular to the red lines. we can write:
AB is perpendicular to OP
BC is perpendicular to OQ
AC is perpendicular to OR
3. Each of the three blue lines PQ, QR and RP subtend an angle at the center O:
♦ PQ subtend ∠POQ
♦ QR subtend ∠QOR
♦ RP subtend ∠ROP
• But PQ = QR = RP (∵ they are sides of an equilateral triangle)
• So the angle subtended by them must be equal
♦ Thus we get: ∠POQ = ∠QOR = ∠ROP
• But the total angle at any point is 360o
♦ We can write: ∠POQ + ∠QOR + ∠ROP = 360o
• So, since the three angles are equal, we get: ∠POQ = ∠QOR = ∠ROP = 360⁄3 = 120o
4. Based on theorem 32.5, we can write:
• ∠A + ∠ROP = 180 ⟹ ∠A+ 120 = 180 ⟹ ∠A = 60
• ∠B + ∠POQ = 180 ⟹ ∠B+ 120 = 180 ⟹ ∠B = 60
• ∠C + ∠ROQ = 180 ⟹ ∠C+ 120 = 180 ⟹ ∠C = 60
• So the three angles of the red triangle are all 60o. This is marked in fig.b. Thus the red triangle ABC is an equilateral triangle.
5. Consider fig.c: AB and AC are two tangents from an exterior point A
(i) P and R are the tangent points. Then by theorem 32.2 that we saw in the previous section, AP = AR
(ii) In the same way, we have: BP = BQ and CQ = CR
(iii) Let: AP = AR = a
BP = BQ = b
CQ = CR = c
(iv) Since the triangle is equilateral, we have: AB = BC = CA ⟹ (a+b) = (b+c) = (c+a)
From (a+b) = (b+c), we get: a = c
From (b+c) = (c+a), we get: a = b
• So we have: a = b = c
6. P is the midpoint of AB (∵ a = b)
R is the midpoint of AC (∵ a = c)
• So the line PR cuts the sides AB and AC at their midpoints
• Then by theorem 18.5, (Division of triangles by parallel lines) we get:
♦ BC is parallel to PR
♦ BC is double the length of PR
In the same way:
♦ AC is parallel to PQ
♦ AC is double the length of PQ
AND
♦ AB is parallel to RQ
♦ AB is double the length of RQ
■ (4) and (6) together gives the answer for part (i)
Part 2:
The steps for construction are given below:
1. Step 1 is to draw an equilateral triangle of side 3 cm. This is shown in fig.32.21(a) below
 |
| Fig.32.22 |
2. Step 2 is to draw the circumcircle of this triangle. The method can be seen here. In this step, we get the position of the circumcentre 'O' also. This is shown in fig.b
3. Step 3 is to join all the three vertices to the centre O. These are the yellow lines in fig.c
Now draw lines perpendicular to the yellow lines through the vertices. These lines (shown in red colour in fig.c) will give the outer triangle
Solved example 32.9
Fig.32.23(a) shows the two tangents (red lines) drawn from an external point. And also the radii (green lines) through the points of contact.
 |
| Fig.32.23 |
(i) Prove that the tangents have the same length
(ii) In fig(b), the external point is joined to the center of the circle (cyan line). Prove that this cyan line bisects the angle between the radii
(iii) Prove that the cyan line bisects the angle between the tangents also
(iv) In fig(c), a chord is drawn between the points of contact (magenta line). prove that the cyan line is the perpendicular bisector of the magenta chord
Solution:
Part (i): The tangents in fig32.23 will be of equal length.We proved it earlier and wrote Theorem 32.2.
Part (ii): We have to prove that the white and grey colored angles in fig.32.24(b) below are equal
 |
| Fig.32.24 |
We have proved it already and wrote the first part of Theorem 32.3
Part (iii): We have to prove that the green and yellow colored angles in fig.32.24(b) above are equal
We have proved it already and wrote the second part of Theorem 32.3
Part (iv): We have to prove that the cyan line is perpendicular to the magenta chord.
♦ And also that it divides the magenta chord into two equal parts as shown in fig.32.24(c) above.
• We proved it in solved example 32.4. See fig.32.13. It is shown again below:
 |
| Fig.32.13 |
• We proved that TO is the perpendicular bisector of PQ
Solved example 32.10
(i) In the fig.32.25(a) below, the two magenta chords are perpendicular to each other.
 |
| Fig.32.25 |
Red colored tangents are drawn at the ends of those chords. Those red tangents meet at four points to give a quadrilateral. Prove that this quadrilateral is cyclic
(ii) What sort of quadrilateral will it be, if one chord is a diameter?
(iii) What sort of quadrilateral will it be, if both chords are diameters?
Solution:
Part (i):
1. Let the magenta chords be PQ and RS. See fig.32.25(b)
2. Red tangents are drawn at P, Q, R and S
• Those tangents meet at four points to give a quadrilateral ABCD
3. The tangent points P, Q, R and S are joined to the centre O by green radial lines
• So, if ∠POR = x, we will get: ∠PAR = (180-x)
• Similarly, if ∠SOQ = y, we will get: ∠SCQ = (180-y)
4. In the quadrilateral ABCD, ∠PAR and ∠SCQ are opposite angles
• Their sum is: [(180-x) + (180-y)] = [360-(x+y)]
5. Consider the arc RP. It is drawn from the ends of perpendicular chords
• Similarly, the arc QS is drawn from the ends of the same perpendicular chords
• If we align the two arcs together, they will form a semi-circle. We proved this as a solved example in a video presentation)
6. The central angles of the arcs are xo and yo. Since they form a semi -circle, we get: (x+y) = 180o
7. So the sum in (4) becomes: [360-(x+y)] = [360-180] = 180
8. That is., sum of opposite angles in the quadrilateral ABCD is 180o. So it is a cyclic quadrilateral. This is shown in fig.32.25(c)
Part (ii):
1. Let the magenta chords be PQ and RS. See fig.32.26(a)
• This time, RS is a diameter
 |
| Fig.32.26 |
2. Red tangents are drawn at P, Q, R and S
• Those tangents meet at four points to give a quadrilateral ABCD
♦ We have to find which type of quadrilateral is ABCD
■ Since PQ and RS are perpendicular chords, the quadrilateral ABCD will be cyclic. This we have proved in part (i)
3. Since RS is a diameter, the tangents at R and S will be parallel.
• That is., the opposite sides AD and BC are parallel
♦ This we proved in solved example 32.5 in the previous section
4. If the opposite are parallel in a quadrilateral, it will be a trapezium
■ If a trapezium is cyclic, it will be an isosceles trapezium.
The proof is given in this video presentation.
Another method:
1. The tangent points P, Q, R and S are joined to the centre O by green radial lines. See fig.32.26(b)
Note that, The radial lines for R and S will be OR and OS themselves. Because RS is a diameter.
• So, if ∠POR = x, we will get: ∠PAR = (180-x)
• Similarly, if ∠SOQ = y, we will get: ∠SCQ = (180-y)
2. Since RS is a diameter, the tangents at R and S will be parallel.
• That is., the opposite sides AD and BC are parallel
♦ This we proved in solved example 32.5 in the previous section
3. So we have two parallel lines cut by a transversal CD
• If the angle at C is (180-y), angle at D will be y (∵ AD and BC are parallel)
See proof in video presentation.
4. But from (6) in part (i), we have: (x+y) = 180
• So we get: y = (180-x)
5. But from (1) above, we have: angle at A = (180-x)
• So angles at A and D are equal
■ If AD and BC are parallel and angles at A and D are equal, it will be an isosceles trapezium.
Part (iii):
1. Let the magenta chords be PQ and RS. See fig.32.27(a)
 |
| Fig.32.27 |
• This time, both PQ and RS are diameters
2. So opposite sides AB and CD will be parallel
The opposite sides AD and BC will also be parallel
3. Since the opposite sides are parallel to the perpendicular lines PQ and RS, we get:
∠A = ∠B = ∠C = ∠D = 90o
• This is shown in fig.32.27(b)
4. Since the corner angles are all 90o, we get:
• Lengths AB = BC = CD = AD
5. From (2), (3) and (4), it is clear that, ABCD is a square.
In the next section, we will see more details about Tangents.
