 Hi, I'm Zor. Welcome to Unisor Education. This lecture is the last theoretical lecture about Apollonius problems. It's basically titled Lines and Circles. It's presented on the website Unisor.com, this one. I suggest you to view this lecture, to watch this lecture from this website because it contains nice notes, as well as just the general functionality of educational process, which includes exams, enrolling, etc. So, as I was saying, this is the last theoretical lecture on Apollonius problems. And before actually going into Lines and Circles, I would like to make a few comments about certain kind of obvious aspects of transformation, which we are talking about. So, inversion. That's the main tool of all these problems encompassed by the name Apollonius. So, this inversion, as I was presenting it, I was presenting it in a relatively straightforward fashion. For instance, this is an inversion circle, the center. And let's say if you have a line here, then you have to draw the perpendicular, draw the image of that point inside, and the image of that line will be a circle, which has this segment as a diameter from a center to a point, which is an image of this point. Now, I did not really go into more complex situations like this, for instance. So, instead of this line, which is completely outside of the circle, what if the line intersects it? What happens in this case? Well, I didn't really touch it because it's really very simple thing. I think it's exactly the same. And as I was saying, the situation is exactly the same. You take this point, you reflect it relative to this inversion circle, and in this case it will be point outside. And again, you construct a circle using this as a diameter, exactly the same type of thing. So, the image of the line is still a circle, and the line intersects the inversion circle. It's just the circle which would be partially in, partially out, as well as this line is partially in, partially out. Now, similarly, if you have a circle which is partially in and partially outside, so this is inversion circle, and you have circle like this, for instance, what's the image? Well, before we were talking about the image of a circle as a circle, we proved it. Well, in this case, again, without the proof, just you have to do exactly the same thing. Have a diameter. Have these two points inverted. So, this one goes into, let's say, this, and this one goes into, let's say, this. Now, these two points which are on the inversion itself, on the inversion circle itself, should stay. And the resulting circle would be this, the image. So, again, a circle goes into a circle without any problems. And what else? Well, if you have two lines, if you have two lines, any lines intersecting each other at one particular point, their image, whatever their image is, would also intersect but the point would be inside. So, if this point belongs to this line, the image of this point belongs to the image of this line, which is way to this, and the image of this line would be something like this. And again, the point would be the same. Intersection would be in one point. Therefore, if you, let's say, have a tangent to a circle, this is an inversion circle. You have a circle and a tangent. So, the circle will be transformed into a circle, in this case, inside. This tangent would be transformed into a circle which goes through the center of the inversion. But my point is there is one common point, the tangential, the point of tangency. So, this circle, whatever it is, would be tangential to this circle. So, this tangential point and this tangential point, they are inversed against each other. Because there is only one intersection. If there is one intersection, it cannot turn into no intersection or two intersections. Obviously, it will be one intersection point among images. And if there is no intersection, there will be no intersection. And here is an interesting example which actually is quite educational as far as what inversion actually is. If you have two parallel lines, well, again, lines do not have anything in common, right? No common points. They are parallel. Which means their image is supposed to be parallel, right? So, the image of this line would be... So, we have to draw it perpendicular. The image of this point would be this. The image of that point would be somewhere here, right? So, this line would transform into this circle which goes through a center of the inversion. And this line will go also to a circle which goes through a center. So, what is it? Is it a common point? No, because if you remember, in the very beginning of explaining what actually inversion is, I said that all outside is converted into the inside except the center. Center is not converted into anything. It's not participating, actually, in the transformation. So, that's why you can say that there is no such point, actually. There is no intersection here. There is no common point, because this is a center and center is completely outside of all the transformations. So, outside, if you just take out this particular point from the inversion circle, there are no intersections between these two images of these two parallel lines. Because there is only one intersection in the center and the center is not part of the picture. There is no such point, actually, if you wish, in this transformation and inversion. So, we still preserve the notion of no intersection, no common points among these parallel lines and there is no common points among its inverted images. Okay, after this preamble, it's time to go to, again, relatively routine lines and circles problems. So, I have... Well, all lines were considered in the first lecture when it's basically a triangle, right? So, you have to inscribe or circumscribe around the triangle. Now, as far as two lines in the circle. Okay, two lines in the circle. I will consider only one particular case. All other cases are similar. Now, the circle is somewhere here. So, we need a circle which would be tangential to all three of them. Okay, how can we accomplish this? Well, if you remember, if, instead of a circle, I have a point it should go through, then I can use this point as a center of inversion, have some kind of a big circle and these two lines would be inverted into circles and this, since it goes through a center, the one which we need to construct would be a line and the line would be tangential to two images of these two lines, which is two circles. So, basically, you would have a circle and a circle and you have to draw a common tangential line which is easy to know how to do that. Now, here, instead of a point, we have a circle but there is a very simple technique which you can actually transform this problem into the problem when the point actually is given. Here is the point. Let's say this is R1, this is R2. What if I will increase the radius R1 by the value of R2? Then it will go through this. It's a concentric circle. Right? Now, this new circle, the bigger one, R1 plus R2, will go through the center of this which is given. Now, what's the difference if I will draw the perpendicular here? That would also be R2, right? And this would be R2. So, if I will shift this line down by R2 and this line this direction by R2 and R2 is known, what do I have? I have a point and two lines and I can now know how to draw a circle which is tangential to two lines and goes through this point, right? That's one of the previous problems. It's a point, line, and line. We know how to do that. And once I've done that, I know the center. And if I know the center, obviously I know the radius because I will subtract from this R2 and draw a new circle which is radius, whatever the radius is. And incidentally, if you have two lines and a point, this problem, how to draw a circle which is passing through this point and tangential? Well, if you remember, we just draw an angle bisector, pick any point, have a circle which is tangential to these two lines by dropping two perpendicular and then we just stretched it. So this point is stretched to this one then the radius will stretch to this one. So basically, it's a proportional thing, right? So you know this, you know this, and you know this, and that's why you can know this. This is the proportionality between these two similar triangles. This and this. Okay. So that's two lines and a circle. Okay, what if you have two circles and one line? So you have a circle which is supposed to be tangential to two circles and a line. Okay. What can we do here? Well, what actually we can do is very similar to the one to the transformation we did before. Let's say the smaller is R1. This is R2 and this is R3. So I will increase the radius of this unknown circle by the smaller one, R1. Now it passes through the center. Now this, it's tangential to the circle of the radius R2 minus R1, right? And this will be tangential to this line which is on the distance R1 and this is R1. So we have reduced our problem to a slightly simpler one. But you have a line, a point, and a circle. Point, line, circle. So you have line, point, and circle. And you have to draw a circle which is this. Now here obviously since we know the point we can use the inversion with the center here. So what we'll do is this is my inversion circle with this center. Now this is supposed to be a straight line, right? This is supposed to be some kind of a circle. And the circle also supposed to be some kind of a circle. So right now the problem after the inversion has been reduced to drawing, constructing a tangent to two given circles. One is image of this line. Well, actually I was wrong. This is not a circle. Since this is a line the image would be not just a circle but a circle which goes through this point. So it doesn't really matter. I mean it's still a circle. But that would be something like this. So this circle is image of this. This circle is image of this line. And this circle would be transforming to this tangent which we have to construct. So again, tangent to two circles. And we know how to do that. Line and two circles. So line and two circles we have reduced to line, circle, and point by reducing the radius. And then use the inversion to solve the problem of line, point, and circle. Now the last one which we did not touch yet that's three circles. Now as I was given this problem construct a circle which is tangential to three given circles. That was actually given me as a true Apollonius problem. I mean all others are kind of derived but this is a regional Apollonius problem. Construct a circle tangential to three given. So what can we do in this particular case? Well actually we can do very simple thing. Again let's take the smallest radius which is this one r1, r2, r3. And we will reduce everything by r1. So this would be point. This would be a circle of the radius r2 minus r1. Right? This is r1. And here also would be this, this is r1, would be a circle r3 minus r1. Now, so what we have now is we have to construct a circle which is passing through this point and tangential to two circles. This, this, and this is the point. So this is a point circle-circle problem which we have considered in the previous lecture when we were talking about points and circles. Again, you just use this point in the center of inversion and these two circles would be converted into, transformed into circles and the circle which we need would be their tangential, tangential line. So basically that's it. That's the solution. So with inversion it seems like very easy. As long as you know how to invert a point to a point, a line to a circle and the circle to a circle, if you know that how to do it and it's very simple actually. Then basically that's it. All problems are resolved. Now the last problem which is kind of a combination we touched before. If you have a point, a line and a circle we already touched it in the previous problem. We have reduced the two lines in this circle. Not two circles in a line. Two circles in a line we have reduced to this one. Alright, so as long as you have a point the easiest way is just to draw a circle around this point as a center and invert everything relative to this circle. So these are all different problems combined together under a general name Apollonius problems. It's constructing a circle which is tangential to three elements and three elements can be anything, points, lines and circles. One more important issue you know these solutions whichever I was suggesting to you I suggested just one solution but in probably almost every case there are more than one solution so if you are really like scientifically approach all these problems you really have to specify what kind of solutions really exist and let me just as an example talk about the last one. So if you have a circle which supposed to be tangential to three given circles how many solutions this problem has? Well, this is one, now this is another right? This is the third one. There are many many circles using the combinatorics we can actually calculate because each circle can be tangential either inside or outside right? This is outside tangency and this is inside tangency so basically I think we have eight different solutions here, two for this, two for this two for this because each one of these three given circles can be either inside or outside touching the circle which we need so eight different solutions and how can we obtain this? So let me just tell it on this particular example if you use first you have to reduce the problem to the problem of two circles at the point by reducing all radiuses to the radius of the smaller one right? So now this is the center so you draw an inversion circle this would be a line, this circle this circle would be another circle now we have two solutions two different tangential lines which are tangential to this one at the same time you can use actually different types of different types of center you can use instead of this center you have three different versions and in all these three different versions the situation would be different obviously and you will have different pictures different circles and different tangents so that's why you have certain freedom of choice and whenever you're exhaust all these variations that would be basically the result of everything so like for instance we are looking for let me just give you an example if you have again this and you have to this get this solution what happens? Now in this case in this case what should you do? well you reduce your radius that would be something like this this and the point so it would be tangential here here and here that's what you're looking for now if you will use this as a center this as a center then this would be circle this would be another circle and you have two solutions here one solution and two solutions one would correspond this circle and another would correspond to this circle right? so basically again all different variations are possible and if you again if you would like to really make a complete solution you really have to think about all the different variations of this problem but to tell the truth the most important I would say interesting part is just to solve it in at least one case like the case when everything is outside I mean that's kind of appealing to my eye and basically everything else follows so you don't really have to struggle with this too much what's important is that there is a very important there is a very powerful tool inversion this is a tool and you have to really master this tool you have to understand how it works and where it can be applied and well with another probably more philosophical side of this is if you will try to solve this problem without the apparatus of inversion it would be probably difficult I didn't think maybe enough for this but right now at the moment inversion is the only method which I know I mean probably there are some others but what's important is that in some cases whenever you are dealing with a difficult problem which you don't really know how to approach using known to you methods what's important is to go completely outside of the box and think about this problem from different perspective I mean somebody invented this particular transformation of inversion I don't even know who by the way but whoever did it it was really a perfect example of thinking outside of the box all these approaches which were used before like for instance you want to inscribe a circle into a triangle well you know how to do it it's angled by sectors etc now these are all classical approaches to construction problem using inversion basically inventing the inversion proving a few theorems that the lines would be converted into circles during this transformation or circles which goes through the center would go into line etc so all these it's a theory the whole apparatus which was built maybe for some other purposes but for this particular purposes of Apollonius problems that was very useful and that happens a lot in science, in theories etc in many cases when you just if you construct some, I know airplane engine with the propeller there is an engine definitely and it can be improved, improved, improved there is certain limit beyond which you cannot go and jets were invented as a completely new type of engine completely outside of the box and they actually made the whole flying significantly faster so you have to think about certain problems which have reached either their level of understanding level of capacity etc you have to think about them completely outside of the box and maybe you will come up with a completely new approach from completely different direction which would be great for this particular kind of thing so this is a perfect example inversion is a perfect example of such an approach in mathematics and whoever invented it is definitely a very talented person thanks very much and good luck