 Next concept is the concept of diameter of a circle Diameter of a circle now normally from our childhood definition We know that diameter is the longest chord right and it passes through the center of the circle But if you go by the actual locus definition of a diameter, so diameter is defined as the locus of Because I'll be using this definition in other conics as well. It's the locus of the middle points of a system of Parallel chords of a circle Okay, so the locus of the middle point of a system of parallel chords becomes the diameter of the circle So basically this definition will be used in all the conics so we'll also have a concept of diameter in case of a Other conics as well. So we'll be using this definition. So let's say this is a circle and we have a Set of parallel chords. So these are let's say parallel chords Okay, and if you join the midpoints of these parallel chords if you join the midpoints of these parallel chords Okay, you get a diameter so this red line that you see that's going to be your diameter of the circle Okay, so now I would request all of you to give me the expression for The equation of a diameter which bisects a system of parallel chords having a slope of M Having a slope of M So please give me the equation of a system of parallel chords of the circle The equation of diameter of a circle x square plus y square is equal to a square which bisects all chords of slope M Which bisects all chords of slope M It should be done by now guys. It's a simple concept Please type done if you're done so that we can start the discussion for it You just say the x plus m y equal to zero. That's absolutely correct So guys, it's pretty simple. Let's say I call the midpoint of this chord as h comma k So I know that equation of this chord is going to be t equal to s1 Correct and T equal to s1 t is going to be xh plus yk minus a square is equal to h square plus k square right Now we also know that the slope of this is M So from this equation we can find the slope the slope will be minus a by b which is equal to minus h by k and This you have to equate it to M Which clearly gives you a relationship between hk and the given information M. So minus h is equal to KM So h plus KM equal to zero now generalize this when you generalize this it becomes x plus My equal to zero so this becomes the equation of the diameter And as you can see it is passing through origin. That means it will always pass through the center of the circle Okay, so now we are moving on to the concept of pole and polar So now we are going to cover up the concept of pole and polar Pole and polar Now the words itself may not be used by J exam, but More importantly, you need to understand how it is connected to the concept of locus How it is connected to the concept of locus that is something which we need to appreciate So let's draw a circle again with center at origin Okay Now within the circle or it may be outside the circle also Let's say there is a point x1 y1. So let's say this is a point x1 comma y1 Again, I'm repeating the point may be within or without the circle Also, it can be outside the circle also. Okay now through this point through this point I draw You know, I start drawing Cods, okay, so let's say I draw a chord like this Okay, and I draw a chord like this So I can keep I keep I can keep on drawing many many cards to these two to this point x1 y1 So infinitely many cards can be drawn now what I'm doing through the cards which I'm drawing I am drawing tangents Okay, so I'm drawing tangents at the end of these quads okay, so So I've drawn a tangent here. I've drawn a tangent here. Hope it doesn't move out of the screen Oh, oh, oh, oh, so I have to be careful Okay, so I'm assuming this is the tangent. Okay, so I assume that this is the tangent now These tangents will start meeting at Certain points whose locus if you trace would be that of a straight line Okay, so these tangents will start meeting on a straight line and That straight line is called the polar For this point which is called the pole Right, so the line is called the polar and The point X1 y1 would be called the pole. So this will be called the pole Okay, and this line will be called the polar Okay, now as a locus if you see this concept is you know the j can frame a question like this through a point x1 y1 You are forming Within the circle you are fine. You are forming Cards and through the ends of the cords you are drawing tangents Find the locus of the meeting point of these tangents, right? So indirectly they are talking about find the equation of the polar Find the equation of the polar, right? Because even for the polar the equation is given by t equal to 0 only Okay, so let's say this is the equation of a circle whose center is at origin Then the equation of the polar Then the equation of the polar will be given by x x1 x x1 Plus y y1 equal to a square only So see for the third time now, I'm using this equation earlier I use it for the tangent at x1 y1 point the other time I use was for the a cord of contact drawn from an external point x1 y1 and This time I'm using it for the for the equation of a polar. So guys the same Expression is now serving three purposes Equation of a tangent equation of a cord of contact and the equation of a polar Now I would request you all first of all to prove this So prove that the locus of h comma k is going to be this expression That is what I'm more interested in rather than telling you the names of pole and polar which probably J We'll never use because they don't want people to come up with all the jargons They want them to be smart enough to derive it or find it out pretty simple It should not take you more than half a minute to solve this MX plus y equal to zero no no you have to prove this push pin there You have to prove that the equation of the polar is this what I've shown you on the screen You have to prove this Okay, guys, it's very simple, you know assume that the other way around assume that from h comma k You are drawing two tangents and these two tangents you know Meet at the circle at this two points Okay, let's say a and b. So a b is the cord of contact and this cord of contact has been satisfied by x1 y1 All right. So first of all the equation of the cord of contact Drawn from h comma k. We all know is again t equal to zero that is xh plus yk Equal to a square. This is the equation of the cord of contact cord of contact That you are drawing from h comma k. Okay, drawn from h comma k, right? Now this cord of contact Let me call it as one. So one is being satisfied by is Satisfied by the point x1 y1 Right because the pole will also lie on the cord of contact So replace your x with x1 Y with y1 Right in equation number one correct right Now if you want to find out the locus of h comma k This is the right relation that you have got between h and k. Now you generalize over here When you generalize it over here, you get x1 x Y1 y equal to a square correct Now it may sound very you know As if I'm scamming it over here But that's the truth if you find the cord of contact from h comma k It is going to be satisfied by x1 y1 and just you generalize your h and k with x and y and Hence we get the equation of the cord of contact. Okay So guys a quick question for you before we move on to the properties of pole and polar Hope there's no doubt with respect to this so my question here is Find the pole find the pole of the line of The line lx plus my plus n equal to 0 With respect to the circle with respect to the circle x square plus y square is equal to a square now This time I have given you the question in in a reverse fashion. I have given you the equation of the Pole are and I'm asking you what is the pole Okay, so just try doing it your answer should be only in terms of l m and a okay No other parameter should be introduced in the answer This is pretty simple guys pretty pretty simple. It should not take you more than 30 seconds Please type done on your screen if you have done it so that I can start the discussion Are in say Ujjah Matli Pushpinder Tapas shares Lalitha Atmesh Nikhil Snigda Nidesh, please feel free to type in your answers Or at least I done so that I can start the process Okay done very easy So guys will assume the point let the point from where you are drawing this polar let the pole be Let the pole be x1 y1 Okay, so we know that if the pole is x1 y1 and this is our circle x square plus y square is equal to a square equation of the polar would be this right Right or in other words, it will be this right and This is same as Lx plus my plus n equal to 0 So these two equations I'm claiming that these are the same equations because as per the question This is the polar right Now we can compare the coefficients. We can compare the coefficients. So it's x1 by L will be equal to Y1 by m and will be equal to minus a square by n, right Which clearly gives you x1 as minus a square L by n and y1 as minus a square m by n Correct. So your pole will be your pole coordinates will be minus a square L by n comma Minus a square m by n right. So this becomes your answer No doubt about that guys, is it clear CLR please type on the screen clear so that I can proceed with the next slide which is on the concept of properties Properties of pole and polar Properties of pole and polar okay, I Start this with the very first question If I say tangent if tangent is a polar, what will be the pole? If the tangent to a circle is a polar, then what should be the pole? Please type it Is my question clear if tangent is treated as a polar then what should be the pole anyone? The point of contact exactly aren't correct. So basically Tangent is the polar Tangent is the polar with Point of contact as the pole point of contact as the pole Okay So this is the very first property you should all know Second property that I'm going to talk about is if O is the center of the circle if O be the center of the circle center of a circle and P be any point P is any point then O P is perpendicular to the polar of P O P is perpendicular to the polar of P a Super super simple property you should be able to prove this within fraction of a second so Let's say this is a point P This is a point P Okay, and there's a polar of P This is the polar of P so it says if you connect O to P if you connect O to P it will fall perpendicular on the polar Can you prove this super simple Very simple, you know that this point Let's say I'm I'm considering it a case of a generic circle a standard circle having the center at origin And let's say the point P is x1 y1 point P is x1 y1 You know that this line is having a equation xx1 plus yy1 equal to a square So the slope of this line is going to be minus x1 by y1 right Whereas the slope of this is the slope of the polar Whereas the slope of OP the slope of OP will be y1 by x1 and If you multiply it slope of the polar into slope of the OP you will always get a minus one Claiming that this angle will always be a 90 degree. Is that fine? Super simple concept moving on to the third property guys Okay, again if O be the center of a circle If O be the center of a circle be the center of a circle and P be any point any point then if OP meets the polar of P in Q then OP dot OQ Or OP into OQ will be square of the radius Will be square of the radius now OP may not meet directly, but if if required or if produced if required Produced if required so OP if produced Meets the polar of P in Q then OP square OP into OQ is radius square So let's quickly construct it. All of you, please prove it at your end also Let's say this is your polar OP into OQ OP into OQ is R square Very very easy guys should not take you more than a minute. So this point is x1 y1 Polar equation you already know so Yeah Let me know if you're done so that I can discuss it Please type done on your chat box so that I know you are done and we can start the discussion for it It's one important thing. I wanted to tell over here It's not necessary that the pole has to be within the circle if the pole is outside the circle You can still form a polar with it and that polar will coincide with the chord of contact So after this problem, I'll probably explain you that as well because many people have this wrong conception or wrong notion that The polar can only be drawn with the pole inside the circle But that is not the case your pole can outside be outside the circle as well Is it done? It's very simple From the origin if I find the distance OQ. I just have to use the distance formula So the distance formula will give me mod a square by under root of x1 square y1 square Correct. This is your OQ and OP is already under root of x1 square plus y1 square Okay, so when you multiply OP into OQ You are left with mod of a square only because this and this will get cancelled off So that's actually square of the radius So now let's take some questions based on pole and polar Yeah before that Let's talk a bit about Polar for those points which are outside the circle. Okay, so if let's say a point is outside the circle x1 y1 Okay, and you're trying to find out the polar for this point again the approach is the same Approach is the same the approach is through this point draw I'm sorry through this point you draw as As many number of lines as you can cutting the circle, so let's say this line this line this line Okay, okay And let's say draw tangent also. So this is the tangent. Let's say this is the tangent. Let's say Okay Now from the meeting point of these lines with the circle draw two tangents Draw two tangents at the meeting point So you would realize that You would realize that The tangents drawn at the extremities of the meeting point of these chords So let's say I'm drawing tangents over here I'm drawing tangent over here and tangent over here. It will meet on the Quad of contact line, so it will meet it over here Okay, so if I draw tangent over here Tangent over here tangent over here it will again meet on the chord of contact line So these the polar would actually become your part of Or it will become an extension of it will become an extension of your chord of contact Code of contact drawn from there code of contact drawn from x1 Y1 point so if you draw a chord of contact from x1 y1 point only extend that line that will start behaving as your polar Right, that's why it has the same equation as the chord of contact t equal to 0 if the point x1 y1 is outside the circle Are you getting this point guys? Okay No question