 Hello all welcome to another session on geometry and in this session we are going to visualize a very common concept and that is on a plane if there is one point then we can have infinitely many circles passing through that point okay so we have been listening to this you know since long long time but in this session we are trying to visualize whether this is actually true and subsequently we are also going to see if there are two points on the same plane then how many circles can be drawn to those two points and finally if there are three then how many can be drawn okay three points then how many circles can be drawn passing through each of the three so let's begin with plotting a point here a so this is point a okay now we are going to draw circle through a so here is another point B and yeah so this is C circle passing through a with center B I'm taking center C and then another circle passing through a so there are definitely two then it's a D circle again passing through a then here E passing through a so all of that and you see there are lots of such circles you can draw which are actually passing through a right so infinitely so this as many points as you can see in the plane you can have those many circles possible through a right I hope this is clear so hence through one point you can draw infinitely many circles okay now let us see what happens when there are two points okay so let us see what happens when there are two points okay let us take a fresh space and we'll do it here so yeah so let's say take two points okay let's say in this case R and S these are the two points through which I am trying to see whether there will be more than one circle so what I'm going to do is I'm going I'm first going to join or finally let's say I am finding the perpendicular bisector of R and S so this line lies or passes through the midpoint of R and S RS okay now if I have a point here T and from this T point if I want to draw a circle let's see what happens so hey so this from this T point to a one circle is passing through both R and S let's take another wow so from you another circle is passing from you a center to another circle is going through R and S and another point V and you can see a third circle is passing to R and S then fourth then fifth then sixth then seventh then eighth and nine and ten like that you can have infinitely many many circles again passing through the two points but the difference between the first case that is when we were trying to draw circles through one point what is the difference guys if you see in the in the previous case you could have taken the center anywhere on the plane but in this case there is a restriction which has got added up the restriction is in this case there will be definitely infinite number of circles passing through the two points R and S but then all the circles will be lying on the perpendicular bisector of segment RS you can see that right let's say if I draw a circle from this point and try to pass through S in this case it passes through S but not through R let's say I take another point here which is on the line RS itself extended but then it passes through R and then if I draw for the same center it passes through S but then there are two different circles so same circle doesn't pass through the two points so this is a learning in the first case you could have center anywhere in the second case you vary the constraint and the constraint is the center would be only on the perpendicular bisector of the two points or through this line segment now let us see what happens if there is a third point added to this scenario so now I have three points one two three okay these are the three points okay now what I'm going to do is I'm going to do again perpendicular bisector of K1 L1 and L1 M1 okay why am I doing this you'll get to know now so if I take this point and one let's say and now try to draw a circle as center okay so if you see it passes through three points right and besides this guy is any other point you take the circle will not pass through the all the three points right so let's try so first of all we'll try random point so let us say this is random point and trying to pass through M1 but then L1 and O1 doesn't pass through let's say now I take a point on the perpendicular bisector here so if you see it definitely passes through M1 and L1 but not with K1 and let's say I take a point here and then yeah so if you see there's no such circle which passes through all these three points there's only one only one circle so hence the constraint we added is earlier we had multiple circles passing through two points but those centers of those circles were on the perpendicular bisector of the given segment K1 L1 right in the previous case it was R and S but in this case if you have another point added then the center is only one so you know center will definitely lie on what the perpendicular bisector but then it has to also lie on the perpendicular bisector of the other chord which is going to be produced like here L1 M1 so hence to two perpendicular bisectors of two non-colinear chords will intersect somewhere and they will intersect exactly at one point so that point is the center now there could be only one point of intersection of two line segments so hence we get only one circle guys okay so whenever there are three points then there will be at least one circle please remember at least it's not that okay there will be cases where there would not be any circle passing through it also no there will be at least one circle and the circle center will be lying on the point of intersection of the perpendicular bisectors of the two chords like that two segments basically so that is the thing now if you increase one more point then here is you know we enter into a realm where there is it's not necessary that there will be a circle passing through all the four points so we can have multiple points and there could be a possibility that there could be one circle passing through this through a let's say you know let's say there are three these are the four points here so if you see I can draw a circle which passes through all of them so let's say this is the center so I can draw a circle which passes through all the four points possible here but it need not these are special case so all the four there are four non-colinear points it need not be that there will be a circle passing through all of them for example if I have points you know you know distributed like that so if I have this kind of a four point set set of four points it's not necessary that there will be a circle which will pass through all the four points you can figure out from the you know the distribution of the point itself here so till three points definitely there is at least one circle which passes through all the three points for three with two points there are infinitely many circles but the constraint is all the centers will lie on the perpendicular bisector of the two line joining the two points and in case of one point there could be infinitely many circles and the centers could be anywhere on that plane so that's what we learned and anything above four and above points there could be a case that there is a circle which is passing through all the points but it's not necessary so that's what we learned right so only in special cases for example in case of a regular you know vertices of a regular polygon there will be a circle passing through all of them so regular hexagon regular pentagon regular you know anything octagon anything you know there will be a circle passing through all the vertices but if it is not regular it need not be there not be a circle which passes through all the points I hope this demonstration made it clear that how circles can be drawn through various points on a plane