Circles

Hello my dear students! Welcome to today's mathematics lesson. I am so happy to be here with you to learn about one of the most beautiful and interesting chapters in geometry - Chapter 9 on Circles.

Now, before we begin, let me ask you something. Have you ever looked at a bangle? Or the wheel of a bicycle? Or perhaps the dial of a clock? What do all these objects have in common? Yes, they are all circular in shape! Circles are everywhere around us, from the plates we eat on to the coins we use. But today, we are going to study circles not just as shapes, but as mathematical objects with wonderful properties and theorems. Are you ready? Let's begin!

First, let us recall what we already know about circles from earlier classes. A circle is the collection of all points in a plane which are equidistant from a fixed point in that plane. That fixed point is called the centre of the circle, and the distance from the centre to any point on the circle is called the radius. We also know about chords, arcs, diameter, secant, and tangent. Today, we will build upon this knowledge and discover some amazing properties of circles.

Let us start with Section 9.1, which is about the angle subtended by a chord at a point.

Now, what do we mean by "angle subtended"? Let me explain with an example. Suppose you have a line segment PQ, and there is a point R which is not on the line containing PQ. If you join PR and QR, then the angle PRQ is called the angle subtended by the line segment PQ at the point R. Simple, isn't it?

Now look at a circle. Consider a chord PQ of a circle with centre O. What is angle POQ? This is the angle subtended by the chord PQ at the centre O. What about angles PRQ and PSQ, where R and S are points on the major and minor arcs respectively? These are the angles subtended by the chord PQ at points R and S on the circle. So now we understand the terminology.

Now, let us investigate the relationship between the size of a chord and the angle it subtends at the centre. If you draw different chords of a circle and measure the angles subtended by them at the centre, what do you observe? You will notice that the longer the chord, the bigger the angle subtended by it at the centre. This makes intuitive sense, doesn't it? A longer chord "covers more" of the circle when seen from the centre.

Now, what happens if we take two equal chords? Will the angles subtended at the centre be the same? Let us find out. Draw two or more equal chords of a circle and measure the angles subtended by them at the centre. You will find that the angles are equal. This leads us to our first theorem.

Theorem 9.1 states: Equal chords of a circle subtend equal angles at the centre.

Let us prove this theorem. Suppose we have two equal chords AB and CD of a circle with centre O. We want to prove that angle AOB is equal to angle COD.

Look at triangles AOB and COD. In these triangles, OA equals OC because they are both radii of the circle. Similarly, OB equals OD because they are also radii. And AB equals CD because that is given to us - the chords are equal.

So in triangles AOB and COD, we have OA equals OC, OB equals OD, and AB equals CD. By the SSS rule of congruence, triangle AOB is congruent to triangle COD. Therefore, angle AOB equals angle COD, because corresponding parts of congruent triangles are equal. We use the abbreviation CPCT for this.

This theorem is very important, and we will use it many times in this chapter. So remember: equal chords mean equal angles at the centre.

Now, what is the converse of this theorem? The converse would be: if two chords of a circle subtend equal angles at the centre, then the chords are equal. Is this true? Let us think about it.

Suppose we have a circle with centre O. We draw an angle AOB at the centre, and let A and B be points on the circle. Now we draw another angle POQ at the centre, equal to angle AOB. If we cut the disc along chords AB and PQ, we will get two segments of the circle. If we put one on top of the other, they will cover each other completely, which means they are congruent. And if they are congruent, then the chords AB and PQ must be equal in length.

This gives us Theorem 9.2: If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

This is the converse of Theorem 9.1. So now we have two-way relationship: equal chords give equal angles, and equal angles give equal chords. This is a very useful property, and we will see many applications of it.

Let me recap what we have learned so far in this section. We learned what is meant by the angle subtended by a chord at a point. We proved that equal chords subtend equal angles at the centre, and we also proved the converse - that equal angles at the centre imply equal chords. These are fundamental results about circles, and they will help us prove many more theorems.

Now, let us move on to Section 9.2, which is about the perpendicular from the centre to a chord.

Let me ask you a question. If I draw a chord of a circle and then draw a line from the centre perpendicular to that chord, what will happen? Will it bisect the chord? Let us find out through an activity.

Draw a circle on a tracing paper. Let O be its centre. Draw a chord AB. Now fold the paper along a line through O so that a portion of the chord falls on the other. Let the crease cut AB at the point M. Then, angle OMA equals angle OMB, which means both are 90 degrees. So OM is perpendicular to AB. Does the point B coincide with A? Yes, it will! So we find that MA equals MB. In other words, the perpendicular from the centre to the chord bisects the chord.

This is exactly what Theorem 9.3 states: The perpendicular from the centre of a circle to a chord bisects the chord.

Let me prove this for you. Consider a circle with centre O and a chord AB. Draw OM perpendicular to AB, where M is the point of intersection. We need to prove that M is the midpoint of AB, which means AM equals BM.

Join OA and OB. In triangles OAM and OBM, we have OA equals OB because they are radii. Angle OMA equals angle OMB because OM is perpendicular to AB. And OM is common to both triangles. Therefore, by the RHS congruence criterion, triangle OAM is congruent to triangle OBM. This gives us AM equals BM, which means M is the midpoint of AB. So the perpendicular from the centre bisects the chord.

Now, what is the converse of this theorem? The converse would be: if a line drawn through the centre of a circle bisects a chord, then it is perpendicular to the chord. Let us state this as Theorem 9.4.

Theorem 9.4: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Is this true? Let us prove it. Let AB be a chord of a circle with centre O. Let M be the midpoint of AB. Join OA and OB. We need to prove that OM is perpendicular to AB.

In triangles OAM and OBM, OA equals OB because they are radii. AM equals BM because M is the midpoint of AB. And OM is common. So by the SSS rule, triangle OAM is congruent to triangle OBM. Therefore, angle OMA equals angle OMB. But these two angles are on a straight line, so they must each be 90 degrees. Hence, OM is perpendicular to AB.

So we have now proved both the theorem and its converse. The perpendicular from the centre bisects the chord, and if a line from the centre bisects a chord, it is perpendicular to the chord. These are two very important results.

Let me recap this section. We learned that the perpendicular from the centre of a circle to a chord bisects the chord, and the line drawn through the centre to bisect a chord is perpendicular to the chord. These are fundamental properties of circles.

Now, let us move on to Section 9.3, which is about equal chords and their distances from the centre.

First, let me explain what we mean by "distance from a point to a line". Consider a line AB and a point P. If you join P to various points on the line, you will get infinitely many line segments. Which one is the distance of the line from the point P? Think about it for a moment. Out of all these line segments, the perpendicular from P to AB will be the shortest. In mathematics, we define this shortest length, the perpendicular from a point to a line, as the distance of the line from the point. If the point lies on the line, the distance is zero.

Now, what about chords of a circle? A circle can have infinitely many chords. If you draw chords of different lengths, what do you notice about their distances from the centre? Let us think about this. The diameter is the longest chord, and since the centre lies on the diameter, its distance from the centre is zero. What about other chords? You will observe that a longer chord is closer to the centre than a shorter chord. This makes sense, doesn't it? The longer the chord, the nearer it is to the centre.

Now, is there a specific relationship between the length of a chord and its distance from the centre? Let us investigate this through an activity.

Draw a circle of any radius on a tracing paper. Draw two equal chords AB and CD of it. From the centre O, draw perpendiculars OM and ON to these chords. Now fold the figure so that D falls on B and C falls on A. You may observe that O lies on the crease and N falls on M. Therefore, OM equals ON. This shows that equal chords are equidistant from the centre.

This gives us Theorem 9.5: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Now, what about the converse? If two chords are equidistant from the centre, are they equal in length? Let us check. Draw a circle with centre O. From the centre O, draw two line segments OL and OM of equal length and lying inside the circle. Then draw chords PQ and RS of the circle perpendicular to OL and OM respectively. Measure the lengths of PQ and RS. Are they different? No, both are equal! This verifies the converse of Theorem 9.6.

Theorem 9.6: Chords equidistant from the centre of a circle are equal in length.

So we have another two-way relationship: equal chords are equidistant from the centre, and chords equidistant from the centre are equal. These are very useful properties.

Now, let me give you an example to illustrate the use of these results.

Example 1: If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.

Let me explain this solution. We are given two chords AB and CD of a circle with centre O, intersecting at a point E. There is a diameter PQ passing through E, such that angle AEQ equals angle DEQ. We need to prove that AB equals CD.

Draw perpendiculars OL and OM on chords AB and CD respectively. Now, in triangle OLE, we have angle LOE equals 180 degrees minus 90 degrees minus angle LEO, which equals 90 degrees minus angle LEO. But angle LEO equals angle AEQ, so angle LOE equals 90 degrees minus angle AEQ. Similarly, angle MOE equals 90 degrees minus angle DEQ. Since angle AEQ equals angle DEQ (given), we have angle LOE equals angle MOE.

Now, in triangles OLE and OME, angle LEO equals angle MEO because they are vertically opposite angles. Angle LOE equals angle MOE (as we just proved). And EO is common. Therefore, triangle OLE is congruent to triangle OME by the AAS criterion. This gives OL equals OM by CPCT.

But we know that if two chords are equidistant from the centre, they are equal. So AB equals CD. And that is what we wanted to prove.

This example shows how we can use the properties we have learned to solve problems. The key idea is to use the relationship between equal chords and their distances from the centre.

Let me recap this section. We learned that equal chords of a circle are equidistant from the centre, and conversely, chords equidistant from the centre are equal. These are powerful results that help us solve many problems about circles.

Now, let us move on to Section 9.4, which is about the angle subtended by an arc of a circle.

First, let us recall what we mean by arcs. The endpoints of a chord, other than the diameter, divide the circle into two arcs - one major and one minor. Now, if we have two equal chords, what can we say about the arcs they subtend? Are they equal in length? They are more than just equal - they are congruent. This means if you take the arc corresponding to one chord and put it on the arc corresponding to the other chord, without bending or twisting, they will completely superimpose on each other.

We can state this as: If two chords of a circle are equal, then their corresponding arcs are congruent, and conversely, if two arcs are congruent, then their corresponding chords are equal.

Now, what is the angle subtended by an arc at the centre? It is defined as the angle subtended by the corresponding chord at the centre. The minor arc subtends the smaller angle, and the major arc subtends the reflex angle (the angle greater than 180 degrees).

Now, here is an important result: Congruent arcs (or equal arcs) of a circle subtend equal angles at the centre. This follows directly from what we learned earlier - equal chords subtend equal angles at the centre, and equal arcs correspond to equal chords.

Now, let us look at one of the most important theorems in this chapter - Theorem 9.7.

Theorem 9.7 states: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

This is a very powerful theorem. Let me explain what it means. Suppose we have an arc PQ of a circle. At the centre O, it subtends angle POQ. Now, take any point A on the remaining part of the circle (the part that does not include the arc PQ). The arc PQ subtends angle PAQ at point A. Then the theorem says that angle POQ equals two times angle PAQ. In other words, the angle at the centre is exactly double the angle at the circumference.

Let us prove this theorem. We need to consider three cases: when the arc PQ is minor, when it is a semicircle, and when it is major.

Let us begin by joining AO and extending it to a point B.

In all cases, angle BOQ equals angle OAQ plus angle AQO, because an exterior angle of a triangle is equal to the sum of the two interior opposite angles.

Also, in triangle OAQ, OA equals OQ because they are radii of a circle. Therefore, angle OAQ equals angle OQA (by the isosceles triangle theorem we learned earlier).

This gives us angle BOQ equals two times angle OAQ. Similarly, angle BOP equals two times angle OAP.

Adding these two equations, we get angle BOP plus angle BOQ equals two times (angle OAP plus angle OAQ).

But angle BOP plus angle BOQ is the same as angle POQ, and angle OAP plus angle OAQ is the same as angle PAQ. So we have proved that angle POQ equals two times angle PAQ.

For the case where PQ is the major arc, the same proof gives us that the reflex angle POQ equals two times angle PAQ.

This theorem has some very important corollaries. First, note that in Theorem 9.7, A can be any point on the remaining part of the circle. So if you take any other point C on the remaining part of the circle, you have angle POQ equals two times angle PCQ, which is also equal to two times angle PAQ. Therefore, angle PCQ equals angle PAQ.

This proves Theorem 9.8: Angles in the same segment of a circle are equal.

Now, what is a segment of a circle? A segment is the region bounded by an arc and the chord joining its endpoints. So if you take any two points on the same side of a chord, the angles they make with the endpoints of the chord are equal. This is a very useful result.

Now, consider the special case when the arc PQ is a semicircle. Then angle POQ is 180 degrees. So angle PAQ equals half of 180 degrees, which is 90 degrees. This gives us another important property: Angle in a semicircle is a right angle.

This is why, if you draw a triangle with the diameter as one side, the angle at the third vertex will always be 90 degrees. This is known as Thales' theorem, and it has been known since ancient times.

Now, what about the converse of Theorem 9.8? Is it true that if a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, then all four points lie on a circle? Yes, this is Theorem 9.9.

Theorem 9.9: If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e., they are concyclic).

Let me explain the proof. Suppose we have a line segment AB which subtends equal angles at two points C and D on the same side of AB. That is, angle ACB equals angle ADB.

To show that A, B, C, and D lie on a circle, let us draw a circle through points A, C, and B. Suppose this circle does not pass through D. Then it will intersect AD (or the extension of AD) at some point, say E.

Now, if points A, C, E, and B lie on a circle, then angle ACB equals angle AEB (because angles in the same segment are equal). But we are given that angle ACB equals angle ADB. Therefore, angle AEB equals angle ADB.

But this is not possible unless E coincides with D. Why? Because in triangle ABD, the exterior angle AEB cannot equal the interior angle ADB unless they are at the same point. Similarly, the other intersection point should also coincide with D. Therefore, D must lie on the circle through A, C, and B. So all four points are concyclic.

This theorem is very useful for proving that four points lie on a circle.

Let me recap what we learned in this section. We learned that equal chords subtend equal arcs, and conversely. We proved that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. We also learned that angles in the same segment are equal, and that the angle in a semicircle is a right angle. Finally, we learned the converse theorem about concyclic points.

Now, let us move on to the final section of this chapter - Section 9.5, which is about cyclic quadrilaterals.

What is a cyclic quadrilateral? A quadrilateral ABCD is called cyclic if all four of its vertices lie on a circle. In other words, it is a quadrilateral inscribed in a circle.

Now, let us investigate the properties of cyclic quadrilaterals. Draw several cyclic quadrilaterals of different shapes and sizes. Measure the opposite angles in each case and add them up. What do you observe?

You will find that in each case, angle A plus angle C equals 180 degrees, and angle B plus angle D equals 180 degrees. This is not a coincidence - it is a fundamental property of cyclic quadrilaterals.

This gives us Theorem 9.10: The sum of either pair of opposite angles of a cyclic quadrilateral is 180 degrees.

Let me prove this theorem. Consider a cyclic quadrilateral ABCD. We need to prove that angle A plus angle C equals 180 degrees, and angle B plus angle D equals 180 degrees.

Join the centre O to the vertices A, B, C, and D. Now, angle A is the angle subtended by arc BCD at point A. Similarly, angle C is the angle subtended by arc BAD at point C.

Now, angle AOB is twice angle ACB (by Theorem 9.7), and angle COD is twice angle CAD. But angle ACB and angle CAD are angles in the same segment (the segment containing arc AB), so they are equal.

Actually, a simpler proof uses the fact that the angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle. But I think you get the idea - the opposite angles of a cyclic quadrilateral sum to 180 degrees.

Now, is the converse true? If the sum of a pair of opposite angles of a quadrilateral is 180 degrees, is the quadrilateral cyclic? Yes! This is Theorem 9.11: If the sum of a pair of opposite angles of a quadrilateral is 180 degrees, the quadrilateral is cyclic.

This is a very important result because it allows us to prove that a quadrilateral is cyclic by checking just one condition - that a pair of opposite angles sum to 180 degrees.

Now, let me give you some examples to illustrate the use of these properties.

Example 2: In the figure, AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Prove that angle AEB equals 60 degrees.

Let me explain the solution. We are given that AB is a diameter, CD equals the radius, and AC and BD intersect at E when extended. We need to prove that angle AEB equals 60 degrees.

Join OC, OD, and BC. Now, since CD equals the radius, triangle ODC is equilateral. Therefore, angle COD equals 60 degrees.

Now, angle CBD equals half of angle COD (by Theorem 9.7, the angle at the centre is double the angle at the circumference). So angle CBD equals 30 degrees.

Also, since AB is a diameter, angle ACB is 90 degrees (angle in a semicircle is a right angle).

Now, in triangle BCE, angle BCE equals 180 degrees minus angle ACB, which is 90 degrees. So angle CEB equals 180 degrees minus angle BCE minus angle CBE, which is 180 degrees minus 90 degrees minus 30 degrees, which equals 60 degrees. But angle CEB is the same as angle AEB. So angle AEB equals 60 degrees. And that is what we wanted to prove.

Example 3: In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If angle DBC equals 55 degrees and angle BAC equals 45 degrees, find angle BCD.

Let me explain the solution. We are given that angle DBC equals 55 degrees and angle BAC equals 45 degrees. We need to find angle BCD.

First, note that angle CAD equals angle DBC because they are angles in the same segment (the segment containing arc CD). So angle CAD equals 55 degrees.

Now, angle DAB equals angle CAD plus angle BAC, which is 55 degrees plus 45 degrees, which equals 100 degrees.

But in a cyclic quadrilateral, opposite angles sum to 180 degrees. So angle DAB plus angle BCD equals 180 degrees. Therefore, angle BCD equals 180 degrees minus 100 degrees, which equals 80 degrees. And that is our answer.

Example 4: Two circles intersect at two points A and B. AD and AC are diameters to the two circles. Prove that B lies on the line segment DC.

Let me explain the solution. We are given two circles intersecting at A and B. AD is a diameter of the first circle, and AC is a diameter of the second circle. We need to prove that B lies on the line segment DC.

Join AB. Now, since AD is a diameter, angle ABD is 90 degrees (angle in a semicircle). Similarly, since AC is a diameter, angle ABC is 90 degrees (angle in a semicircle).

Therefore, angle ABD plus angle ABC equals 90 degrees plus 90 degrees, which equals 180 degrees. This means that D, B, and C are collinear. In other words, B lies on the line segment DC. And that is what we wanted to prove.

Example 5: Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic.

Let me explain the solution. We are given a quadrilateral ABCD. Let AH, BF, CF, and DH be the angle bisectors of angles A, B, C, and D respectively. These bisectors form a quadrilateral EFGH. We need to prove that EFGH is cyclic.

Now, angle FEH equals angle AEB, which equals 180 degrees minus angle EAB minus angle EBA (by the angle sum property of a triangle). But angle EAB is half of angle A, and angle EBA is half of angle B. So angle FEH equals 180 degrees minus half of (angle A plus angle B).

Similarly, angle FGH equals angle CGD, which equals 180 degrees minus half of (angle C plus angle D).

Therefore, angle FEH plus angle FGH equals 180 degrees minus half of (angle A plus angle B) plus 180 degrees minus half of (angle C plus angle D), which equals 360 degrees minus half of (angle A plus angle B plus angle C plus angle D).

But the sum of the angles of any quadrilateral is 360 degrees. So angle A plus angle B plus angle C plus angle D equals 360 degrees.

Therefore, angle FEH plus angle FGH equals 360 degrees minus half of 360 degrees, which equals 360 degrees minus 180 degrees, which equals 180 degrees.

So the sum of angle FEH and angle FGH is 180 degrees. Therefore, by Theorem 9.11, the quadrilateral EFGH is cyclic. And that is what we wanted to prove.

This example shows how we can use the properties of cyclic quadrilaterals to solve more complex problems.

Now, we have covered all the major concepts in this chapter. Let me summarize everything we have learned.

In this chapter on Circles, we started by understanding what is meant by the angle subtended by a chord at a point. We proved that equal chords of a circle subtend equal angles at the centre, and conversely, equal angles at the centre imply equal chords.

We then learned about the perpendicular from the centre to a chord. We proved that the perpendicular from the centre of a circle to a chord bisects the chord, and the line drawn through the centre to bisect a chord is perpendicular to the chord.

Next, we learned about the relationship between equal chords and their distances from the centre. We proved that equal chords of a circle are equidistant from the centre, and conversely, chords equidistant from the centre are equal in length.

After that, we studied the angle subtended by an arc of a circle. We proved that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. We also learned that angles in the same segment of a circle are equal, and that the angle in a semicircle is a right angle. We proved the converse about concyclic points as well.

Finally, we studied cyclic quadrilaterals. We learned that the sum of either pair of opposite angles of a cyclic quadrilateral is 180 degrees, and the converse - if the sum of a pair of opposite angles of a quadrilateral is 180 degrees, the quadrilateral is cyclic.

These are the key concepts and theorems from Chapter 9 on Circles. These properties and theorems are not just theoretical - they have many practical applications in geometry and in solving real-world problems.

Remember, geometry is a beautiful subject, and circles are one of the most elegant figures in mathematics. The properties we have learned today have been known for thousands of years and are still used in many areas of mathematics and science.

I hope you enjoyed this lesson and understood all the concepts clearly. Thank you for your attention, and keep practicing these theorems and properties. They will help you solve many problems in geometry. Good luck, and see you in the next lesson!