Circles

Namaste students, welcome to today's mathematics class. I am so happy to see you all here, ready to learn something new and interesting. Today, we are going to study Chapter 10 from your NCERT textbook, and that is about Circles. Now, you have already studied about circles in your previous classes, especially in Class IX, where you learned that a circle is a collection of all points in a plane which are at a constant distance from a fixed point, and that fixed point is called the centre. The constant distance is called the radius. You also learned about terms like chord, segment, sector, and arc. So, we already have some foundation about circles. Now, in this chapter, we are going to take our understanding of circles to a much deeper level. We are going to study about tangents to a circle, and this is a very important topic that will help you in solving many geometric problems.

So, let us begin by understanding what happens when we have a circle and a line in a plane. Consider a circle and a line PQ. Now, there can be three different possibilities regarding how this line and the circle are positioned with respect to each other. First possibility is that the line PQ and the circle have no common point at all. In this case, we call the line a non-intersecting line with respect to the circle. The second possibility is that the line PQ and the circle have two common points. In this case, we call the line PQ a secant of the circle. And the third possibility is that there is only one point which is common to both the line and the circle. In this case, we call the line a tangent to the circle. These are the three basic positions that a line can have with respect to a circle. Can there be any other position? Think about it. No, there cannot be any other position. A line can either miss the circle completely, or cut through it at two points, or just touch it at exactly one point.

Now, let me give you some real-life examples to understand these concepts better. You must have seen a pulley fitted over a well, which is used for drawing water from the well. If you look at the rope on both sides of the pulley, and if you consider each side of the rope as a ray, it is like a tangent to the circle representing the pulley. The rope just touches the pulley at one point and then goes away. That is exactly what a tangent does. Another example is a bicycle or a cart moving on the road. Look at the wheels of a bicycle. All the spokes of a wheel are along the radii of the circle. Now, when the bicycle moves on the ground, the wheel touches the ground at exactly one point. That line along which the bicycle moves is actually a tangent to the circle representing the wheel. So, tangents are all around us in real life.

Now, let us understand what exactly a tangent is. A tangent to a circle is a line that intersects the circle at only one point. This is the definition that you must remember. The point where the tangent touches the circle is called the point of contact. In our Fig. 10.1(iii), the point A is the point of contact, and we say that the tangent touches the circle at point A.

Now, the next question that arises is: does a tangent always exist at every point on the circle? How can we be sure that for any point on the circle, we can draw a tangent? Let us perform some activities to understand this better.

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 this system on a table and gently rotate the wire AB about the point P to get different positions. In various positions, the wire intersects the circular wire at P and at another point Q1 or Q2 or Q3, and so on. But in one particular position, you will see that it will intersect the circle only at the point P. This shows that a tangent exists at the point P of the circle. If you rotate it further, you will see that in all other positions, the wire will intersect the circle at P and at another point. So, what we observe is that there is only one tangent at a point of the circle.

While doing this activity, you must have noticed something interesting. As the position of AB moves towards the position where it becomes a tangent, the common point Q1 of the line AB and the circle comes nearer and nearer to P. Ultimately, it coincides with P. Similarly, if you rotate AB in the opposite direction, the common point R3 comes closer and closer to P and ultimately coincides with P. This shows us something very important: the tangent to a circle is a special case of the secant, when the two endpoints of its corresponding chord coincide. What does this mean? When we have a secant line that cuts the circle at two points, those two points define a chord. Now, if we move one point of the chord closer and closer to the other point, the chord becomes shorter and shorter. When the two points finally coincide, the chord becomes just a point, and the secant becomes a tangent. This is a beautiful relationship between a secant and a tangent.

Activity 2: On a paper, draw a circle and a secant PQ of the circle. Now, 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. That means the two points of intersection of the line and the circle are coming closer and closer to each other. 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. These are the tangents to the circle parallel to the given secant PQ. This activity also helps us see that there cannot be more than two tangents parallel to a given secant.

So, students, we have understood that a tangent exists at every point on the circle, and there is only one tangent at each point. Now, let us look at a very important property of the tangent. We have already seen in the bicycle wheel example that the radius through the point of contact appears to be at right angles to the tangent. Let us prove this property mathematically.

Theorem 10.1 states: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Let us prove this theorem. 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. Now, the point Q must lie outside the circle. Why? Because if Q lies inside the circle, then the line XY would become a secant, not a tangent. Since Q is outside the circle, the distance OQ is longer than the radius OP. That is, OQ is greater than OP. Now, this happens for every point on the line XY except the point P. So, OP is the shortest distance from O to all the points on XY. And from geometry, we know that the shortest distance from a point to a line is the perpendicular distance. Therefore, OP is perpendicular to XY. This is Theorem 10.1.

This is a very important theorem, and you must remember it: The tangent at any point of a circle is perpendicular to the radius through the point of contact. This also confirms that at any point on a circle, there can be one and only one tangent. The line containing the radius through the point of contact is also sometimes called the normal to the circle at that point.

Now, let us discuss how many tangents can be drawn from a point on or outside a circle. This is a very interesting question.

Consider a point P inside the circle. Can you draw a tangent to the circle through this point? If you try, you will find that all the lines through this point will intersect the circle in two points. It is not possible to draw any tangent to a circle through a point inside it. This is Case 1.

Now, consider a point P on the circle. We have already seen that there is exactly one tangent to the circle at such a point. This is Case 2.

Finally, consider a point P outside the circle. If you try to draw tangents to the circle from this point, you will find that you can draw exactly two tangents to the circle through this point. This is Case 3. In this case, the points where the tangents touch the circle are called the points of contact. Let us call them T1 and T2. The length of the segment from the external point P to the point of contact is called the length of the tangent from that point to the circle. So, PT1 and PT2 are the lengths of the tangents from P to the circle.

Now, here is a very interesting property. If you measure PT1 and PT2, you will find that they are equal. Is this always true? Yes, it is! Let us prove this.

Theorem 10.2 states: The lengths of tangents drawn from an external point to a circle are equal.

Let us prove this. We are given a circle with centre O, a point P lying outside the circle, and two tangents PQ and PR drawn from P to the circle, where Q and R are the points of contact. We need to prove that PQ is equal to PR.

Join OP, OQ, and OR. Now, by Theorem 10.1, we know that the tangent is perpendicular to the radius at the point of contact. So, angle OQP is a right angle, and angle ORP is also a right angle. So, we have two right triangles, triangle OQP and triangle ORP. In these triangles, OQ equals OR because they are radii of the same circle. OP is common to both triangles. So, we have two sides equal in both triangles, and the hypotenuse is common. Therefore, by the RHS congruence criterion, triangle OQP is congruent to triangle ORP. Since the triangles are congruent, the corresponding sides are equal. Hence, PQ equals PR. This is what we wanted to prove.

We can also prove this using the Pythagoras theorem. In right triangle OQP, by Pythagoras theorem, we have PQ squared equals OP squared minus OQ squared. Similarly, in right triangle ORP, PR squared equals OP squared minus OR squared. But since OQ equals OR, we get PQ squared equals PR squared, which gives us PQ equals PR.

There is another interesting observation from this theorem. Since triangle OQP is congruent to triangle ORP, we also get that angle OPQ equals angle OPR. This means that OP is the angle bisector of angle QPR. In other words, the centre of the circle lies on the bisector of the angle between the two tangents drawn from an external point.

Now, let us look at some examples to understand these concepts better.

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.

Two concentric circles means they have the same centre. Let the larger circle be C1 and the smaller circle be C2, both with centre O. We are given a chord AB of the larger circle C1 which touches the smaller circle C2 at the point P. We need to prove that AP equals BP, that is, the chord is bisected at the point of contact.

Join OP. Now, AB is a chord of the larger circle, and it touches the smaller circle at P. So, for the smaller circle, AB is a tangent at point P, and OP is its radius. Therefore, by Theorem 10.1, OP is perpendicular to AB. Now, we have a chord AB of the larger circle, and a perpendicular from the centre O to the chord AB. We know that the perpendicular from the centre of a circle to a chord bisects the chord. Therefore, OP bisects the chord AB. Hence, AP equals BP. This is what we wanted to prove.

Example 2: Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that angle PTQ is equal to twice angle OPQ.

We are given a circle with centre O, an external point T, and two tangents TP and TQ to the circle, where P and Q are the points of contact. We need to prove that angle PTQ equals two times angle OPQ.

Let angle PTQ be equal to theta. Now, by Theorem 10.2, we know that TP equals TQ. So, triangle TPQ is an isosceles triangle. Therefore, angle TPQ equals angle TQP, and each of these angles is equal to half of 180 degrees minus theta, which is 90 degrees minus half theta.

Also, by Theorem 10.1, angle OPT is 90 degrees because OP is the radius and TP is the tangent at point P. So, angle OPQ equals angle OPT minus angle TPQ, which is 90 degrees minus (90 degrees minus half theta), which equals half theta. So, angle OPQ equals half of angle PTQ, or in other words, angle PTQ equals two times angle OPQ. This is exactly what we needed to prove.

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. Find the length TP.

We are given a circle with radius 5 cm. There is a chord PQ of length 8 cm. The tangents at P and Q intersect at point T. We need to find the length of TP.

Join OT. Let OT intersect PQ at point R. Now, since TP and TQ are tangents from point T, we know that TP equals TQ, so triangle TPQ is isosceles. Also, TO is the angle bisector of angle PTQ. Therefore, TO is perpendicular to PQ, and it bisects PQ. So, OR is perpendicular to PQ, and PR equals RQ, which is 4 cm each.

Now, in triangle OPR, we have OP equals 5 cm, and PR equals 4 cm. Since angle ORP is a right angle (because OR is perpendicular to PQ), we can find OR using Pythagoras theorem. OR squared equals OP squared minus PR squared, which is 5 squared minus 4 squared, which is 25 minus 16, which is 9. So, OR equals 3 cm.

Now, consider triangle TPR. We have angle TRP equals 90 degrees. Also, angle TPR plus angle RPO equals 90 degrees, and angle TPR plus angle PTR also equals 90 degrees. So, angle RPO equals angle PTR. Therefore, triangle TRP is similar to triangle PRO by AA similarity.

So, we have TP divided by PO equals RP divided by RO. That is, TP divided by 5 equals 4 divided by 3. So, TP equals 20/3 cm, which is approximately 6.67 cm.

We can also find TP using the Pythagoras theorem in another way, but the result is the same.

So, students, in this chapter, we have learned several important concepts. Let me summarize everything that we have studied in this chapter.

First, we learned about the three possible positions of a line with respect to a circle: non-intersecting line, secant, and tangent. A tangent is a line that intersects the circle at only one point.

We learned that a tangent to a circle is perpendicular to the radius through the point of contact. This is Theorem 10.1, and it is a very important result.

We also learned about the number of tangents from a point. There is no tangent to a circle passing through a point inside the circle. There is exactly one tangent to a circle passing through a point on the circle. And there are exactly two tangents to a circle passing through a point outside the circle.

We learned Theorem 10.2, which states that the lengths of tangents drawn from an external point to a circle are equal. This is a very useful result and is used in many proofs and problems.

We also learned some interesting properties, like the centre of the circle lies on the angle bisector of the angle between the two tangents drawn from an external point.

We saw examples of how to apply these theorems to solve problems, like proving that a chord of a larger circle touching a smaller concentric circle is bisected at the point of contact, proving the relationship between angles in a circle with tangents, and finding the length of a tangent segment.

So, students, these are all the important concepts from Chapter 10 on Circles. Make sure you understand these theorems and their proofs well. Practice the examples again and again. Thank you for your attention, and I will see you in the next class. Good luck with your studies!