 Hello, welcome to the Koenig section YouTube live session today So those who have just joined in Please type in your name in the chat box so that I know who all are attending this session All right guys, so today's session We are going to start with the fact that why at all Koenig sections are known by the name Koenig sections okay, so Koenig section is basically Given the name because they have been obtained by the section of a right circular double cone with a plane Right, so these Koenig sections they are obtained by They are obtained by the cross section of a right circular double cone a right circular double cone by a plane Okay, so depending upon how these planes or how this plane passes through the right circular double cone Different Koenig sections are generated. Now. What is the right circular double cone a cone which has got? Let me explain this to you first a cone is basically Right the right circular double cone is basically a structure like this where this is called the axis of the cone So this line is called the axis of the cone. Okay, and these two lines are known as the generators These two lines are known as the generators When the axis of the cone makes 90 degree with the base of the each cone We call it as a right circular double cone. Okay, so I will now show you How basically these Koenig sections are generated? Okay, so let us focus on Let's focus on this structure So I can as you can see on your screen. Okay, so basically this is nothing but a right circular double cone Okay, so I've shown it in a 3d structure now this right circular double cone if I cut it by a plane which is perpendicular to the axis Right as you can see here I've cut it by a plane which is perpendicular to the axis the cross section area from the top You can see is that of a circle Okay, so the cross section area that you can see is that of a circle Okay, if you cut it by such a cone or Such a plane which is making some angle Okay Then you realize that you start getting different types of Koenig's Okay, in this case you end up getting a hyperbola Right, so these basics are all cleared in your mind and If there is any question regarding the basic of creating a Koenig section, please do ask in fact, I will quickly brush you through So when you cut this by a plane When you cut this by a plane Whose angle with the axis let's say is Let's say I call the semi vertical angle of this right circular cone as as alpha and let's say this angle is theta Okay, the angle of between the plane and the Axis is theta if theta is 90 degree it generates a circle Okay, and if this plane passes through the meeting point of these right circular double cone the circle would be a point circle Okay, if this theta is somewhere between 0 to somewhere between Alpha to 90 degree Alpha to 90 degree it will create an ellipse It will generate an ellipse Again here. I'm assuming that it is not passing through the meeting point of the generators okay, if Theta is exactly equal to alpha That means the angle made by the plane is parallel to either of these generators The cross-sectional area left would be that of a parabola. Okay, and if theta lies between 0 to alpha Then it would end up cutting both the Naps these are called naps Okay, both the naps of this right circular double cone and You will obtain a hyperbola in that case If it passes through the meeting point if it passes through P if the conic section passes through Or if the plane passes through P if the plane happens to pass through the point P Then the next we call the conic generated as a degenerate conic Okay, and pair of straight lines is an example of a degenerate conic So first I will spend some time on the degenerate conic in order to explain you What is the general form of the equation of a conic section? So let's talk about pair of straight lines briefly What's a pair of straight lines? pair of straight lines is basically a kind of a degenerate cone where you obtain the Equation of this pair of straight line by multiplying two lines Okay So if you start by multiplying two such lines which are passing through Let's say origin y equal to m1x and y equal to m2x Okay, if I ask you these are the lines which make that pair of straight lines Then the combined equation of the pair of straight lines would be y minus m1x times y minus m2x equal to 0 right, so this is the Equation of a pair of straight lines which is passing through origin Which is passing through or intersecting each other at origin So both the straight lines are passing through the same point origin. Okay, if you clearly look at this expression It's very important for you to appreciate that This expression will be a homogeneous Expression in x and y okay, so we call this as a Homogeneous function or homogeneous equation in two variables This is going to be a homogeneous equation in two variables of degree two of Degree two so what is a Homogeneous equation a homogeneous equation is basically a equation where you would realize all the variables are having the same degree for example x square Y square x y they are all having the same degree. Okay, so any second degree Equation which is homogeneous in nature in two variables that would always represent a pair of straight lines passing through origin This is very very important so always remember this Any equation of this nature a x square plus 2 h x y plus b y square Will always represent a pair of straight lines Passing through origin Now it is not necessary that both the lines would be real in nature It may happen that you may also end up getting non real lines as well, right? You may end up getting imaginary lines as well that depends upon some relationship between h a and b that I will discuss with you later on Okay, now taking a clue from this You would realize that any conic that you see is going to be having a general formula or the general equation as a x square plus 2 h x y plus b y square plus 2 g x Plus 2 f y plus c equal to 0 Right, this is the general structure of any conic section that you would come across Right, which is called a second-degree general equation of a conic section. So even your straight lines as you can see here The the homogeneous its equation itself is basically is a second-degree equation Even circle parabola ellipse will have these six terms appearing not on it's not necessary that all of them will appear together It may happen that some of them may be absent Now, how why is this a second-degree equation? Guys, if you if you see a very specific case from the example which I cited a little while ago the equation of the equation of if you see the equation of Cone I've taken passing through the origin as z square equal to x square plus y square, right? So this is the equation of the cone okay, and If you take any line any plane, okay, let's say I take this plane x plus y plus z equal to 1 Okay, and you try to Eliminate your z from these two equations So from the plane equation and the cone equation if you try to eliminate your z Okay, you would realize that Let's say this was the equation of the cone Okay, and let's say this is the equation of the Plane any plane. Okay, so I'm taking just a simple Example of a cone and a plane if you try to eliminate your z that means replace your z with 1 minus x minus y in this You yourself would see that When you square such terms You'll end up getting some terms in x square Some terms in y square You'll see some term with x y You'll see some term in x you'll see some term in y and you'll see a constant, isn't it? So that is what we are trying to state in the previous slide that Any conic section would have This kind of a structure. Sorry this kind of a structure a x square Plus 2 h x y plus d y square plus 2 g x plus 2 f y plus c equal to 0 Right so far any questions with respect to this Okay, so this is the general structure of any conic section Now depending upon your values of a b and a b h g f and c this conic section can represent various things Right, so now we'll discuss the condition to represent various conics Condition to represent various conics The first condition that I'm going to talk about that's the very important condition if a x square plus b y square plus 2 h x y plus 2 g x plus 2 f y Plus c equal to 0 has to represent a pair of straight lines If it has to represent a pair of straight lines then an expression delta which is given as a b c plus 2 f g h minus a f square minus b g square minus c s square will be equal to 0 that is In a simpler form the determinant made by these terms Will be equal to 0 Right now first of all I will prove how these condition comes so let us prove how this condition comes Okay So let us start with the fact that let a x square Plus b y square plus 2 h x y Plus 2 g x plus 2 f y plus c equal to 0 represent a pair of straight lines Okay, let's say this represent a pair of straight lines Of course since this is not a homogeneous equation that means Their point of intersection of these lines is not origin because had it been origin These three terms would have been absent right So let's say these pair of straight lines This pair of straight lines intersect each other at Let's say they intersect each other at x1 y1 and we are assuming that these lines are also not parallel to each other. Okay Then do one thing then apply the concept of shifting of origin shifting of origin to x1 y1 That means I'm assuming the point of intersection now to be my new origin okay, so when we do that all of you remember the concept of Shifting of origin where you replace your x with Where you replace your x with capital x plus x1 and y with capital y plus y1 So this was the concept which we had discussed in the beginning of the coordinate geometry chapter in class 11 So when you do these replacements in your equation given to you you get something like this a x plus x1 square Okay, b y plus y1 square plus 2h x plus x1 y plus y1 2g x plus x1 2f y plus y1 Plus c equal to 0 okay, and If you simplify this you are going to see something like this a x square by square 2h x y and Apart from that you will see terms like 2a x1. I'm collecting terms having capital X in it so 2a x1 I will see then from here. I will get h y1 and I will also get term like 2g Similarly, if I collect terms containing capital Y I Will get 2b y1. I will get 2h x1 and I will get 2f term correct and If I collect the constant terms, I will get a x1 square b y1 square 2h x1 y1 2g x1 2f y1 plus c equal to 0 Okay Now all of you please listen to this very very carefully if You want this pair of straight lines to pass through origin It is passing through origin Then it is very clear that it has to be a homogeneous equation And if it has to be a homogeneous equation that means this term this term and This term should be 0 Remember, I discussed with you Any homogeneous equation of Second-degree is basically representing a pair of straight line passing through origin and Since I am shifting my origin to the point of intersection x1 y1. It means that I am assuming that These terms are going to be 0 This term is going to be 0 This term is going to be 0 and this term is going to be 0. Is that clear? Because if these three become 0, I will only end up getting this term Which is a second-degree term and hence it will represent a pair of straight lines passing through origin so this implies We have a x1 H y1 plus g equal to 0 We have this is x1. Yeah, this is H x1 plus b y1 plus f equal to 0 and Guys, if you note your expression very very carefully If you note these expressions very very carefully you'd realize that you can write this as x1 into a x1 plus H y1 plus g plus This expression I am simplifying plus y1 into H x1 plus b y1 plus f and The remaining terms would be g x1 f y1 plus c So what I did is I took an x common from this term and One of these terms I split it open as H x1 y1 plus H x1 y1 So I took a x1 common from this term and I took x1 common from One of these terms. So this also a spitter spit it up as g x1 plus g x1 This also a spit up as f y1 plus f y1 So one of the terms a x1 h y1 and g has come from taking this term this term and this term Similarly for y1 if I take y1 common from this term This term and this term I get this term. I will get this term, okay, and the remaining terms that is G x1 f y1 and c I am writing it over here Is that clear any question regarding this so far now from 1 and 2 we already know that these two are Zero this is zero. This is zero Correct. That means I get a third equation from here. That is g x1 f y1 plus c also equal to zero So I get three equations altogether Like this. So this is my third equation now using these three equations I would now eliminate x1 y1 Okay, so from this you try to eliminate x1 y1 Okay, and how do I eliminate it if you remember the trivial solution concept of a System of homogeneous equation or a system of linear homogeneous equation the determinant If this determinant is equal to zero Okay, then this system of equation. So guys what I'm trying to say is You know your H, you know your x1 y1 and 1 right? They are not going to be a trivial solution over here Okay, so if this system of equation doesn't have any trivial solution We can say that the determinant formed by the coefficients Okay, so basically let me treat one over here if nothing is there So the coefficient of these variables would form a determinant, which is equal to zero, right? And if you expand this you are going to end up getting the desired result, which is a b c plus 2 f gh Minus af square Minus bg square minus c s square equal to zero. Is that fine? Is that clear? So this is the condition which we write in a very short form as delta Okay, so if delta is zero your second-degree equation will definitely represent a pair of straight lines Now what about the rest of the conics? What about the rest of the conics like circle parabola ellipse hyperbola? What can you comment about all these things? So for that we have to understand the locus definition of all the conics Okay So what is a conic section? The conic section is basically locus of a point which moves in a plane in such a way in Such a way that the ratio of its distance from a fixed point perpendicular distance from a fixed Line is always a constant Okay, and this constant is known by the name of eccentricity and this fixed point it can be fixed points as well They are known by the name of focus and this fixed line is known by the name of directx. That is what we already know okay now there is This word eccentricity which comes up and this eccentricity actually helps us to differentiate between different different conics Okay, if e is equal to 1 It would represent a parabola okay, if e is Between 0 and 1 it will represent an ellipse if e is greater than 1 it will represent a hyperbola if e is equal to 0 it would represent a circle and if e is tending to infinity a very very large number Then it will become a pair of straight lines. So hyperbola is a Limiting case when e tends to infinity your hyperbola will start behaving as a pair of straight lines Now, how is this e connected to your? Second-degree equation. How is my e connected to the second-degree equation that we had learned a little while ago? So we realize that if e is equal to 1 if e is equal to 1 That means if it represents a parabola Then these coefficients of these this equation would actually satisfy the condition h square is equal to AB Okay, now a question will arise in your mind. How? Right, we have been studying this when h square is equal to AB in this equation It will represent a parabola. So, how is this e equal to 1 linked to this h square equal to AB? Okay, for that I would write the equation of a general conic Okay, so let's say the fixed point focus Let's say it is a point which is I can say alpha comma beta Or let's say h comma k and let's say the fixed line or what we call as the directrix So let the directrix be Lx plus my plus n equal to 0 now from the locus definition From the locus definition of a conic section This is going to represent a conic for me that is the distance from the fixed point is e times the distance of The point x comma y from this fixed line right I should not take h and k because h and k has already been used. Let's take alpha beta for the sake of More clarity Let's say alpha beta Okay, here also we'll take an alpha beta Okay Now let us square both the sides if you square both the sides you get x minus alpha square y minus beta square is equal to e square by l square m square l square m square And lx plus my plus n whole square Okay, let's cross multiply l square m square times x minus alpha square y minus beta square is equal to e square And we have lx plus my plus n whole square Okay, now We all see that a is basically when you say h square is equal to ab h is basically the Half of the coefficient of x y half the coefficient of x y a is basically the coefficient of x square b is basically the coefficient of y square, right? So let us focus only on these terms. So if I ask you what is a Your answer will be l square plus m square That is what we get from multiplying of this term with this term Okay, and from here I will get minus e square l square That is I get l square 1 minus e square plus m square correct If I ask you what is the coefficient b from this equation You would say coefficient of y square, which is this term when it multiplies to this term That is l square m square Okay, and you will have a minus m square coming up from here m square e square So that is going to be l square m square 1 minus e square Correct and what is your H term Okay H term you would say can only be obtained from this expression Correct that will be e square e square l m That would be e square l m correct So when these when you square this term you will get 2 l m x y Yes or no And so the coefficient is 2 e square l m half it you will get h Is that clear? Now what I will do is I will do A b minus h square Okay So a b is basically product of these two Minus square of this term Now product of these two if you see it will become l 4 1 minus e square m 4 1 minus e square And we will get something like this l square m square 1 plus 1 minus e square whole square correct So when these two terms multiply, I'll get the first term when these two terms multiply I'll get the second term and when you cross multiply these I will get this term. Okay minus e to the power 4 l square m square. Let's not forget this term as well So here I will get 1 minus e square l to the power 4 m to the power 4 Okay, and I will also get terms like l square m square 1 plus 1 plus e to the power 4 minus 2 e square Okay, and of course minus e to the power 4 from here. Correct This term this term gets cancelled So if you want you can take 1 minus e square common from both the terms So you'll end up getting something like this minus 2 l square m square okay So this becomes 2 This becomes minus 2 e square. So 2 I have taken out common Okay, so if I take a 2 out common I will be left with this Sorry, it will be plus in between It will be plus in between. Okay Now this is clearly 1 minus e square times l square plus m square whole square. Isn't it? Okay Now guys listen to this explanation very very carefully. In fact, I will take you to the next page So, we know a b a b minus h square is 1 minus e square times l square plus m square whole square So, let me take you to the next page Now you realize that when e is 1 When e is 1 What will happen to the right side of the expression? What will happen to the right side of this expression? You will say that the right-hand side of the expression will become 1 minus 1 times l square plus m square whole square That is equal to 0 Isn't it and that is the reason why we get h square is equal to a b For a parabola getting this point Yes or no When e is between 0 and 1 what happens? When e is between 0 and 1 what happens? You would realize that 1 minus e square Will be a positive term see This term would be a positive term And this term is anyways a positive term This term is anyways a positive term That means your a b minus h square would be positive in nature That means your h square will be less than a b and this is the condition for it to represent A ellipse This was the condition for it to represent a parabola Are you getting this point? So, how is this e related to h square minus a b is what I am trying to explain you over here If e is greater than 1 What happens? What happens? This term over here will become a negative term Whereas this term is going to be a positive term. So negative into positive will be a negative term That means a b minus h square will be less than 0 That means if h square is greater than a b it would represent a hyperbola It would represent a hyperbola Is the point clear? So how these conic sections are coming it is clear to you and if e is equal to 0 what is happening? If e is equal to 0 you would realize that your a b minus h square will be as good as l square plus m square whole square correct If you go back If you go back You would realize that when e is 1 your a itself is When e is equal to 0 When e is equal to 0 a itself becomes l square plus m square And b also becomes l square plus m square Which clearly indicates that in case of a circle In case of a circle you get an additional condition that Your H should be 0 this will become 0 correct Yes or no Your H will become 0 because your e is 0 and your a will be equal to b Do you remember you had learned this condition in class 11? That for a circle we have an additional condition that a should be equal to b And H would become 0 Are you getting this point? Now for all these cases which I have discussed so far Your delta should not be 0 Your delta should not be 0 for all these conditions Else it will become pair of straight lines Is that fine? So guys, this is the basic introduction of a conic section And today I'm going to start my discussion with a circle In fact, I'm quickly going to Take you through j e main and j e advance concepts But at the same time I would quickly revise whatever you have learned in circles so far without you know much delay So let us start our discussion with the first conic which is the simplest of all which is the circle Okay I'm not going to take a lot of time For revision just a few things that I want you to all know first You should know the equation of the circle in various forms Okay, so we'll start with equation in various forms first one is the center radius form the center radius form which is quite Familiar to all of you guys So if we have a circle Okay, if you have a circle Whose center is at h comma k and whose radius is going to be a units Its equation is given as x minus h whole square y minus k whole square equal to a square No doubt about it very simple concept Second is the equation of a circle in parametric form For the same circle, which I have mentioned right now over here For the same circle if you write the equation in parametric form We write it as x equal to h plus a cos theta And y is equal to k plus a sin theta Here theta is a parameter Theta is a parameter if you try to eliminate this parameter You will end up getting the center radius form of the equation of the circle Okay Quickly moving on this is all known to you So I'm not going to spend too much time on it Third is The general form The general form of the equation of a circle As we have already seen in case of a general second degree equation A and B should be same So what I did is I assumed them to be one or you say you can divide throughout with A or B And this becomes the general form of the equation of a circle Where the center of the circle is given as minus g minus f And the radius is given as Under root of g square plus f square minus c So this is also known to you guys no need to waste too much time on this The fourth form is the diametrical form Diametrical form Of the equation of a circle so whenever a circle is given to you Whose extremities of the diameter? So let's say this is the diameter So let's say extremities of the diameter are given to us as x1 y1 and x2 y2 Then we write the equation of this circle as x minus x1 x minus x2 Plus y minus y1 times y minus y2 equal to zero Okay So nothing hard and fast about this. This was all known to you fourth is The equation of a circle equation of a circle passing through passing through three non-colinear points three non-colinear points Very important normally people tend to forget this It is in a form of a determinant Okay Where the first of the determinant is x square plus y square plus x y1 Second row is x1 y1 square passing through three non-colinear points. So let the point be x1 y1 x2 y2 and x3 y3 x3 y3 So it will be a determinant given by x square plus y square x y1 x1 square plus y1 square x1 y1 1 x2 square plus y2 square x2 y2 1 And x3 square plus y3 square x3 y3 1 Now guys, how does this equation come about this equation comes in a very simple way So let the equation be x square plus y square plus 2g So i'm going to prove this quickly 2 fy plus c equal to zero let this be the equation of the circle Okay And since it satisfies these three points x1 y1 So x1 y1 will satisfy this equation x2 y2 will also satisfy this condition And x3 y3 will also satisfy this condition Okay now All I need to do is I need to eliminate Eliminate gf and c from this equation Okay Now again treat these numbers as coefficients Okay These numbers as coefficients Of gf and c and 1 Okay Treat them as coefficients So these are all coefficients of g f and c and 1 And the determinant formed by them should be equal to zero Okay So the determinant formed by all these coefficients should be zero because you know Your gf c etc and 1 cannot be trivial Right So treat this in light of a trivial solution of system of homogeneous equation You directly get this expression I would request you to remember this expression because it proves To save a lot of time You cannot afford to sit and derive this in the examination hall Right and it's very easy to use this formula Okay So these are the various forms of the equation of a circle Now we are moving to the next concept intercepts made by The x and the y axis is In fact, you can say intercept made on The axis is By a circle So let's say There is a circle like this Okay This is my y axis and this is my x axis So basically if I want to know how much intercept has been cut by this circle, let's say it's x square plus y square plus 2 gx Plus 2 f y plus c equal to zero on the x axis Okay That means I need to know what is the length ab Please note that ab is not the diameter. It is the intercept Cut on the x axis So how can I find ab? Very simple. You'll say put y as zero Right when you put y as zero You get x square plus 2 gx plus c equal to zero And we know that this point Will be something like x 1 comma zero this point will be something like x 2 comma zero And what I need is what I need is mod of x 1 minus x 2 correct So how will I get that very simple? You know its roots are going to be x 1 and x 2 Correct. So x 1 plus x 2 will be minus 2 g x 1 into x 2 will be c So we know the formula x 1 minus x 2 whole square Is x 1 plus x 2 whole square minus 4 x 1 x 2 That is going to be 4 g square minus 4 c So x 1 minus x 2 mod will be under root of this which is going to be 2 under root g square minus c So remember this result. It's very very helpful in many cases However, you can also find it out by using the fact that you know the distance of the origin from the x axis that is nothing but the Ordinate of the center and you know the radius So you can always find this length and double it up. That's another way But it's very important to understand or look at this concept from quadratic equation point of view as well Similarly the intercept cut on the y axis. Let me call it as cd Okay intercept cut on the y axis will be 2 under root f square minus c So please remember this result cd is 2 under root f square minus c And ab is 2 under root g square minus c Please remember this result as well moving on Next concept is position of a point position of a point With respect to a circle With respect to a circle So if you have been given any circle whose equation is x square plus y square plus 2 g x plus 2 f y plus c equal to zero The first thing that I want to discuss over here guys This term henceforth would be referred to by the name s Okay s stands for second degree equation So this is a second degree term. So I will represent this term by s Okay So this expression is x This expression is called s Okay Now when I substitute let's say I talk about this point x1 y1 being inside the circle Okay, let's say it is at position a When you substitute x1 y1 into In place of x and y I would call that expression as s1 Okay So if the point is located within the circle that means in case Where the point is at a You realize that your s1 would be negative that is The value of this term that would be obtained by substituting such a point which is within the circle would be a negative number Okay, if it lies on the circle that is your case b Of course, you know the point is going to satisfy the equation. So s1 is going to be zero And if it is outside the circle, let's say at point c So for case c it would be greater than zero Okay By the way s1 is called the power of the point With respect to the circle With respect to the circle So we say power of a point if it is negative at the point lies within the circle If the power of a point is equal to zero it lies on the circle And if power of the point is greater than zero it lies outside the circle And one important thing I wanted to discuss over here is that If the point actually lies outside the circle Let's say I take a small case over here Let's say this is the point x1 y1 Okay So when you draw a tangent from this line on to this circle, so let's say I call it as pt Then length of pt is going to be under root of s1. Please remember this Okay, so if you square root the power of a point You will end up getting the length of the tangent drawn from the external point on to the circle and as you can clearly see If the point is within the length of the tangent will come out to be imaginary And if the point is on the circle, it will come out to be zero And if it is outside the circle, it will come out to be positive Next is the concept of intersection Of a line And a circle Intersection of a line and a circle So let me draw a circle quickly over here. So let's say this is a circle And let's say we have a line Then this line And this circle will interact in three ways One is this Another is like this And another is like this Okay So there can be three situations The line may cut the circle as in a or touch the circle as in b Or not even touch the circle as in c Okay So now let the equation of I'll take a very simple case. Let the line be y is equal to mx plus c And let the circle be the standard case Of a circle That means having its center at origin and radius as a Okay Now in order to know the condition for a b and c what I will do I will simultaneously solve the equation of this line with the circle So I will simultaneously solve this So I would replace y with mx plus c in the second equation Which will end up giving me this quadratic equation Okay And if this quadratic equation has So let's say I call this one So if one has real roots Okay, if one has real roots Then it will lead to the case a Okay if one has Sorry real and distinct roots. I should say real and distinct roots. It will lead to condition a If one has real and equal roots It would lead to condition b that is the line would touch the circle and if one has Imaginary roots it will lead to your condition c correct So real root means b square minus 4 ac should be greater than zero that will give you 4 m square c square minus 4 1 plus m square c square Minus a square would be greater than zero Right So if you simplify this you will end up getting a square Let me drop the factor of 4 from everywhere So if you drop the factor of 4 you'll get m square c square minus This greater than zero expand it Okay So these two terms will get cancelled So you'll get c square lesser than a square plus a square m square That is Your c square will be less than a square 1 plus m square for you to have Real and distinct roots. So this is your case a So this becomes your case a Correct Similarly Without wasting much time. I can say for case b Your c square will become equal to a square 1 plus m square Okay, this condition is of importance to us because we call this as the condition of tangency We call this as a condition of tangency For a line y equal to mx plus c to be tangent to The standard case of a circle x square plus y square is equal to a square and finally You can say situation number c Where c square will be greater than a square 1 plus m square Is that clear? So now we'll discuss more about the second condition over here. That is the condition of tangency So again, let me first make a circle over here So this is the standard case of a circle Now this is called the point of contact p is called the point of contact Now if I mention that this line has a slope of m This line has a slope of m And it happens to be tangent to this circle Okay Then what could be the possible equations of this line? So you can say that there can be two situations for A line having a slope of m to be tangent to this given Circle one can be a line as shown to you and another can be a line like this Right both these lines will have the same slope of m correct So how do we get these two equations? How do we get the equation of line? root Plus minus a under root 1 plus m square Okay That means the equation of L1 and L2 Can be written as y equal to mx plus minus a under root 1 plus m square Okay, guys, this is called the Slope form of the equation of a tangent slope form of equation of a tangent to this circle now Can you guys help me to get the point of contact? How do I find p and q that is the point of contact of this line with the circle? Guys, is it audible? Can you see the screen? Am I visible to you guys? Tapas, Atmaish, Matli Can you see the screen? Oh Let me check Can you see it now? Oh voice is audible, but the screen is black Still not visible But it says you are Sharing the screen already Now is it visible? Some problem has occurred Now is it visible? Just Let me try Please let me know when it is visible Is it visible now? Let me let me drop this and send you a new link, okay? Let me just end this broadcast