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Class X - Maths: Tangents and Secants to a Circle
One Word Answer Questions:
Q) What is radius?
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Q) What is center?
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Q) What is tangent?
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Q) What is tangent to a circle?
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Q) Definition of circle?
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Short Answer Questions:
Q) Prove the theorem "The tangent at any point of a circle is perpendicular to the radius through the point of contact"?
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Q) Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact?
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Q) The lengths of tangents drawn from an external point to a circle are equal. Prove.
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Long Answer Questions:
Q) Tangent to a Circle explain the Activity 1.
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Q) Tangent to a Circle explain the Activity 2.
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Q) Tangent to a Circle explain the Activity 3.
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Q) Explain the Tangent to a Circle?
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  • A tangent to a circle is a line that intersects the circle at only one point.
  • The tangent to a circle is a special case of the secant, when the two end points of its corresponding chord coincide.
  • Tangent is the secant when both of the end points of the corresponding chord coincide.
  • radius:A straight line from the centre to the circumference of a circle or sphere.
  • centre:the point that is equally distant from every point on the circumference of a circle or sphere.
  • A circle is a collection of all points in a plane

Contents

  • Introduction
  • Tangent to a Circle
  • Number of Tangents from a Point on a Circle

Introduction
  • A circle is a collection of all points in a plane which are at a constant distance (radius) from a fixed point (centre). You have also studied various terms related to a circle like chord, segment, sector, arc etc. Let us now examine the different situations that can arise when a circle and a line are given in a plane.
  • radius:A straight line from the centre to the circumference of a circle or sphere.
  • centre:the point that is equally distant from every point on the circumference of a circle or sphere.
  • Circle is the locus of points equidistant from a given point, the center of the circle.
  • The common distance from the center of the circle to its points is called radius. Thus a circle is completely defined by its center (O) and radius (R)
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Fig. 10.1

In Fig. 10.1 (i), the line PQ and the circle have no common point. In this case, PQ is called a non-intersecting line with respect to the circle.
Fig. 10.1(ii), there are two common points A and B that the line PQ and the circle have. In this case, we call the line PQ a secant of the circle.
Fig. 10.1(iii), there is only one point A which is common to the line PQ and the circle. In this case, the line is called a Tangent to the circle.

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Fig. 10.2

You might have seen a pulley fitted over a well which is used in taking out water from the well.
Look at Fig. 10.2. Here the rope on both sides of the pulley, if considered as a ray, is like a tangent to the circle representing the pulley.

  • Is there any position of the line with respect to the circle other than the types given above? You can see that there cannot be any other type of position of the line with respect to the circle.
  • In this chapter, we will study about the existence of the tangents to a circle and also study some of their properties.

Tangent to a Circle

Tangent to a Circle

In the previous section, you have seen that a tangent to a circle is a line that intersects the circle at only one point.

To understand the existence of the tangent to a circle at a point, let us perform the following activities:

Activity 1:

Take a circular wire and attach a straight wire AB at a point P of the circular wire so that it can rotate about the point P in a plane. Put the system on a table and gently rotate the wire AB about the point P to get different positions of the straight wire [see Fig. 10.3(i)].

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Fig.10.3(i)

In various positions, the wire intersects the circular wire at P and at another point Q1 or Q2 or Q3, etc. In one position, you will see that it will intersect the circle at the point P only (see position A'B' of AB). This shows that a tangent exists at the point P of the circle.

  • On rotating further, you can observe that in all other positions of AB, it will intersect the circle at P and at another point, say R1 or R2 or R3, etc.
  • So, you can observe that there is only one tangent at a point of the circle. Fig. 10.3 (i)
  • While doing activity above, you must have observed that as the position AB moves towards the position A' B', the common point, say Q1, of the line AB and the circle gradually comes nearer and nearer to the common point P. Ultimately, it coincides with the point P in the position A'B' of A''B''.
  • Again note, what happens if 'AB' is rotated rightwards about P? The common point R3 gradually comes nearer and nearer to P and ultimately coincides with P.
The tangent to a circle is a special case of the secant, when the two end points of its corresponding chord coincide.


Activity 2 :

On a paper, draw a circle and a secant PQ of the circle. Draw various lines parallel to the secant on both sides of it.
You will find that after some steps, the length of the chord cut by the lines will gradually decrease, i.e., the two points of intersection of the line and the circle are coming closer and closer [see Fig. 10.3(ii)].

  • In one case, it becomes zero on one side of the secant and in another case, it becomes zero on the other side of the secant.
  • See the positions P′Q′ and P′′Q′′of the secant in Fig. 10.3 (ii).
  • These are the tangents to the circle parallel to the given secant Fig. 10.3 (ii)PQ.
  • This also helps you to see that there cannot be more than two tangents parallel to a given secant.
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Fig. 10.3 (ii)

This activity also establishes, what you must have observed, while doing Activity 1, namely, a tangent is the secant when both of the end points of the corresponding chord coincide.
The common point of the tangent and the circle is called the point of contact [the point A in Fig. 10.1 (iii)]and the tangent is said to touch the circle at the common point.

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Fig. 10.4

Now look around you. Have you seen a bicycle or a cart moving? Look at its wheels.
All the spokes of a wheel are along its radii. Now note the position of the wheel with respect to its movement on the ground.
Do you see any tangent anywhere? (See Fig. 10.4).
In fact, the wheel moves along a line which is a tangent to the circle representing the wheel.
Also, notice that in all positions, the radius through the point of contact with the ground appears to be at right angles to the tangent (see Fig. 10.4). We shall Fig. 10.4 now prove this property of the tangent


Theorem 10.1 :

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Proof :
We are given a circle with centre O and a tangent XY to the circle at a point P. We need to prove that OP is perpendicular to XY.
Take a point Q on XY other than P and join OQ (see Fig. 10.5).
The point Q must lie outside the circle. (Why? Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius OP of the circle. That is,OQ > OP.
Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY. (as shown in Theorem A1.7.)

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Fig. 10.5
Remarks :
  1. By theorem above, we can also conclude that at any point on a circle there can be one and only one tangent.
  2. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point.
Number of Tangents from a Point on a Circle

  • To get an idea of the number of tangents from a point on a circle, let us perform the following activity:

Activity 3 :

Draw a circle on a paper. Take a point P inside it.
Can you draw a tangent to the circle through this point? You will find that all the lines through this point intersect the circle in two points.
So, it is not possible to draw any tangent to a circle through a point inside it.
Next take a point P on the circle and draw tangents through this point.
You have already observed that there is only one tangent to the circle at such a point [see Fig. 10.6 (ii)].
Finally, take a point P outside the circle and try to draw tangents to the circle from this point.
What do you observe?
You will find that you can draw exactly two tangents to the circle through this point [see Fig. 10.6 (iii)].
We can summarise these facts as follows:
Case 1 : There is no tangent to a circle passing through a point lying inside the circle.

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Fig. 10.6 (i)
Case 2 : There is one and only one tangent to a circle passing through a point lying on the circle.
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Fig. 10.6 (ii)
Case 3 : There are exactly two tangents to a circle through a point lying outside the circle.
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Fig. 10.6 (iii)
In Fig. 10.6(iii), T1and T2 are the points of contact of the tangents PT1 and PT2 respectively.
The length of the segment of the tangent from the external point P and the point of contact with the circle is called the length of the tangent from the point P to the circle.
Note that in Fig. 10.6 (iii), PT1 and PT2 are the lengths of the tangents from P to the circle. The lengths `PT_1` and `PT_2` have a common property. Can you find this? Measure PT1 and PT2. Are these equal? In fact, this is always so. Let us give a proof of this fact in the following theorem.


Theorem 10.2 :

The lengths of tangents drawn from an external point to a circle are equal.
Proof : We are given a circle with centre O,a point P lying outside the circle and two tangents

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Fig. 10.7

PQ, PR on the circle from P (see Fig. 10.7).
We are required to prove that PQ = PR.
For this, we join OP, OQ and OR. Then`/_ OQP` and` /_ ORP` are right angles, because these are angles between the radii and tangents,and according to Theorem 10.1 they are right angles.
Now in right triangles OQP and ORP,
OQ = OR
OP = OP
Therefore, OQP `~=` ORP
This gives PQ = PR

Remarks :
  1. The theorem can also be proved by using the Pythagoras Theorem as follows:
    `PQ^2 = OP^2 – OQ^2 = OP^2 – OR^2 = PR^2 (As OQ = OR)` which gives PQ = PR.
  2. Note also that `/_OPQ = /_ OPR`. Therefore, OP is the angle bisector of `/_QPR`, i.e., the centre lies on the bisector of the angle between the two tangents.


Let us take some examples.

Example 1 : Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Solution :
We are given two concentric circles C1 and C2 with centre O and a chord AB of the larger circle C1 which touches the smaller circle C2 at the point P (see Fig. 10.8). We need to prove that AP = BP.

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Fig. 10.8

Let us join OP. Then, AB is a tangent to `C_2 `at P and OP is its radius.
Therefore, by Theorem
`OP _|_ AB`
Now AB is a chord of the circle `C_1` and `OP _|_ AB`. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,
i.e., AP = BP


Example 2:

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that `/_ PTQ = 2 /_ OPQ`.

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Fig. 10.9

Solution :
We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (see Fig. 10.9). We need to prove that
`/_ PTQ = 2 /_OPQ`
Let `/_ PTQ = theta `
Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle.
Therefore, `/_ TPQ = /_ TQP `=`1/2` (180° − `theta`)
=90° − `(1/2)theta`
Also, by Theorem 10.1, `/_ OPT `= 90°
So,` /_ OPQ = /_ OPT – /_ TPQ `
= 90° −[90° − `(1/2)theta`]
= 90° −[90° +`(1/2)theta`]
= `1/2 theta`
=`1/2 /_ PTQ`
This gives `/_ PTQ = 2 /_ OPQ`


Example 3 :

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 Fig. 10.10). Find the length TP.

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Fig. 10.10

Solution:
Join OT. Let it intersect PQ at the point R. Then TPQ is isosceles and TO is the angle bisector of `/_ PTQ`. So, `OT _|_ PQ` and
therefore, OT bisects PQ which gives PR = RQ = 4 cm.
Also, OR = `sqrt(OP^2 − PR^2)`
= `sqrt(5^2 − 4^2 )` cm
= 3 cm .

Now, ` /_ TPR +/_ RPO =` 90° `=/_ TPR + /_ PTR` (Why?)
So, `/_ RPO =/_ PTR`
Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.
This gives `(TP)/(PO) =( RP)/(RO) `,
i.e., `(TP)/5 =4/3`
or `TP = (20)/3cm`.
Note : TP can also be found by using the Pythagoras Theorem, as follows:
Let TP = x and TR = y. Then
`TP^2=TR^2+QR^2`(Taking right PRT)
`x^2 = y^2 + 16` (1)
`TP^2+OP^2=OT^2`(Taking right OPT)[here OT=OR+RT]
`x^2 + 5^2 = (y + 3)^2` (2)
Subtracting (1) from (2), we get
`25 = 6y – 7 or y = (32)/6 = (16)/3 `
Therefore, `x^2 =(16/3)^2+16`
=`((16)/9)(16+9)`
=`(16 * 25)/9 [From (1)]`
or ` x = (20)/3`

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