 Hello, welcome all getting from Centrum Academy. So welcome to the Koenig section session three Those who have joined in the session just now I would request you to type your name in your names in the chat box so that I know who all are attending the session So guys are in this session. We are going to resume from where we left off in the last class We were doing Koenig sections and we were into the circles part and we had already discussed a lot of concepts So today's session I would be starting with the concept of chord bisected at a given point So the first concept that we're going to look into today is chord bisected at a given point chord bisected at a given point So let's say I have a circle So let's say this is a circle and I've been given a chord on this circle Whose midpoint is known to me as x1 y1 Let's say the chord AB and there's a midpoint given to me P Which is x1 comma y1 Okay Let the equation of the circle be x square plus y square equal to a square So I'm considering the circle to be centered at the origin and having a radius of a So this equation of the chord AB Equation of the chord AB Would be given by the expression t equal to s1 T equal to s1. So you already know what is the expression t? So those who have forgotten about that. Let me quickly recap t is basically nothing but x x1 plus y y1 minus a square This is called the expression t for this circle Right, if you circle changes to the general form that is x square plus y square plus 2gx plus 2f5 plus c equal to 0 In that case it would become xx1 yy1 gx plus x1 fy plus y1 Plus c Plus c okay And what is s1? s1 is nothing but when you substitute x1 y1 in place of x and y This is going to be your s1 So guys all of you quickly prove this that the equation of this chord will be given by x x1 y plus yy1 minus a square is equal to x1 square plus y1 square minus a square in other words it will be xx1 Plus yy1 equal to x1 square plus y1 square Let's try proving this first It's very very simple You just have to use your basic idea of forming the equation of a line When you know a point on it and when you know the slope of that line. So guys, let me tell you if you draw if you draw a perpendicular from the center on this chord Right, it's going to meet the chord at x1 y1. So this angle here would be 90 degree Okay, so we know the slope of OP slope of OP is going to be y1 by x1 Right, so we also know the slope of AB So AB slope would be minus x1 by y1 right So once we know the slope and we know one of the points on AB It's very easy to construct the equation of the line as y minus y1 is equal to minus x1 by y1 x minus x1 Which gives you the equation as yy1 minus y1 square minus xx1 plus x1 square Which is nothing but xx1 plus yy1 is equal to x1 square plus y1 square Right, so this is for a circle whose center is at origin However, you can use the same formula this formula can be used universally across all the types of circles So if we're using for a general circle in case of a general form of the equation of a circle It just becomes xx1 yy1 plus gx plus x1 plus fy plus y1 equal to x1 square y1 square plus 2gx1 plus 2fy1 Right, c and c will get cancelled off Is that fine? So no question with respect to the chord bisected at a given point So let us quickly take a question on this so that you understand the concept well So the question is find the equation of the final equation Find the equation of the chord find the equation of the chord of the circle x square plus y square minus 6x plus 10y minus 9 equal to 0 Which is bisected at Which is bisected at Which is bisected at minus 2 comma 4 which is bisected at minus 2 comma 4 So please do this and type in your answer in the chat box So all you need to do is just replace your x square with xx1 x This is your x1. This is your y1 xx1. It becomes minus 2x yy1 means for y Okay, and minus 6x could be written as minus 3 x Plus x1 Okay, then why could be written as 5 y plus y1 Okay, y plus y1 and This would be equal to x square which is equal to minus 2 square y square Which is equal to 4 square minus 6x Plus 10 y and minus 9 minus 9 anyways gets cancelled from both the places. So I'll not bother writing it So let's collate the terms. We'll have a minus 5x if I'm not wrong We'll have a 9y We'll have a 9y Right, and let's see the constants that come up Atmesh, please check you're working. I think some somewhere some mistake has happened So work the constant terms would be 6 plus 20 and on this side will be 4 plus 16 which is 20 and 12 Plus 40 right So 20 and 20 gets cancelled. So finally your answer should be minus 5x plus 9 y Minus 46 minus 46 equal to 0. So this should be your answer, right? All right, so it was an easy question to start with let's move on to the next question on the same concept Now this is a locus question find the locus of the middle points of Cods Middle points of Cods of The circle x square plus y square is equal to a square Which Which Subtents a right angle Which subtends a right angle at the point c comma 0 at the point c comma 0 To find the locus of the middle points of the Cods of the circle x square plus y square equal to a square Which subtend a right angle at the point c comma 0 Any idea guys how to proceed with this Hi Bush bender All right guys, so let's look into the situation now So we have a circle I'm really drawing a big circle so that all of you can see the points which are marked inside it and We have a chord like this Okay, so let's say there's a chord like this and at the same time. Let me draw the Coordinate accesses Now this is the midpoint H comma k whose locus we desire so we call it as h comma k Okay, so let me call this point as n Let me call this as a let me call this as b and There is a point c comma 0 somewhere over here. Okay, so let's say I point this as c comma 0 point Okay, let me call it by the name of P Okay Now guys a couple of things that we need to note over here is that when you drop a perpendicular from the center Right, it's going to hit the chord at h comma k Okay, so this is going to be 90 degree no doubt about it Okay, and secondly the question also says that The chord will subtend a right angle at point P right, so if you connect these two It's subtending a right angle at the point P So this angle is also a right angle as per the as per the question Okay, now if it's up to answer right angle, okay, can we all say that a Line connecting end to P so let me draw Let me draw a line connecting end to P with a pink color. Okay, so let's say End to P. So please don't get confused. This is your center. Oh Okay, this is point P. This is point a and this is point B. Okay Now when you see In this figure you realize that angle a P o Sorry a P B angle a P B is the right angle Right Yes, I know What does it mean? It means that you can actually draw a circle which passes through a B and P and A so you can actually draw a circle like this. Okay, so I'm not Drawing the full circle. So a circle can be drawn through a P B such that Such that a B is the diameter of that circle because we know that diameter Subtends a right angle at the vertex, correct So you can always pass a circle So you can always construct a circle through Through a P B such that a B is the diameter of the circle correct if a B is the diameter Okay. Now, can I say one thing that? Since n is the midpoint of a B n is going to be the center of the circle and will be the center of the circle Because n is the midpoint of a B is the midpoint of a B, right? Now if that is the case Then a n would be equal to n B will be equal to n P because they will all be the radius of the circle So n P would be equal to n B is equal to n P Right Yes or no Okay now Focus on focus on the triangle focus on the triangle O N B Focus on the triangle O N B In triangle O N B Right guys, I can say that Let me connect it by a straight line over here In triangle O N B. I can say that O B square will be equal to O N square will be plus Plus N B square. Yes or no, right? so From these two set of information that we have one is this and another is this I Think I should be able to find the locus of the point N Okay, so from the first equation I can say N B N B square Will be equal to N P square Correct, so from this relation I can say N B square is equal to N P square and What is N B square? We don't know right, but can we find it from here? So I can say N B square is equal to OB square minus O N square Right, right, so I can say N B square from equation number two is OB square minus O N square So this I'm replacing with OB square minus O N square Now OB square is nothing but the radius of the circle correct and O N square will be nothing but H square plus K square, right? And N P square will be nothing but H minus C whole square plus K square Right and this is what I wanted. I wanted to get a relationship between H K C and A Right, so now let's simplify this. So we'll have a square is Equal to let's expand this term H square minus 2 CH plus C square Plus K square plus H square plus K square Okay, so if you simplify this further, you will get a square minus C square is equal to 2 H square plus K square minus 2 CH okay, now the obvious step is the step of Generalization, so let's generalize this and when you generalize this you get a square minus C square is twice of X square plus Y square minus 2 C X So this is going to be your desired locus of the midpoint of the chord AB So guys again, I'll repeat the process what I did was First of all, I figured out from this figure itself that you can actually Since it is uptending a right angle at P. You can actually draw a circle with AB as the diameter And since N is the midpoint of AB A N and NB become the radius and so will be NP So they will be all equal that is your condition number one and Then we use Pythagoras theorem on triangle O NB. That is OB square is equal to ON square plus NB square now from the first condition I can write NB square is NP square and NB square itself can be obtained as OB square minus ON square and All these values that I have that is OB ON NP I can always write in terms of HK and C etc Right any questions, please ask me. Is it clear type C LR if it is clear to you If it is clear then I can move on to the next concept