 Hi, I'm Zor. Welcome to Inezor Education. We continue talking about Apollonius problems. Now, the previous two lectures were very important. The first lecture actually talked about how to solve simple Apollonius problems. No, not really supposed to be even called Apollonius. However, they are kind of prerequisites for the real Apollonius problems when the circles are involved. Now, the second lecture was about inversion as a transformation of the points on the plane, which helps to transform circles into lines to help us to resolve these problems. Let me just make a very very brief reminder. The first lecture was about constructing a circle which is either passing through points or is tangential to lines. So, we had four different problems. We had three points and the way how you draw a circle to pass through all these points is to have a segment by sectors. This and this, which is perpendicular to the line which connects these two. And that's why these lines, these segments would be the same. So, that's when three points are. When you have two points and a line, two points and a line, and you have to construct a circle which is tangential to the line and passing through these two points. Now, what I have suggested was, if you remember, there is a theorem about this. Let's say this is P, A, B and this is C, that P A times P B is equal to P C square. So, basically that is a sufficient condition to find the P C. So, if you connect A and B and continue until the intersection at point P with this line, you will find point P and then you calculate P C as P A times P B. And that is, can be done by, for instance, constructing a right triangle with this B and this will be P C. If this is A P and this is P B, then this will be square, P C square will be A P times P B. Now, the third problem was two lines and one point. And two lines and one point. So, what I have suggested was, just construct any circle which is tangential to these two lines and stretch this circle into this. If this is all, this is P prime, let's say, and this is P, then basically O P prime divided by O P is the same as O M prime divided by O M, because these triangles are similar. So, basically knowing O P prime and O P, you can end O M, because this is just a circle which we have chosen. We can find the center. And the third, sorry, the fourth problem was when all three are lines. So, you have three lines here and to inscribe a circle into a triangle, you basically use the angle bisectors. Now, obviously, you can do it externally as well, something like this, when you use these bisectors. All right. So, these are simple problems. Inversion is not involved. There are no circles given only lines and points. Now, then I have introduced the concept of inversion, the most important part of it. Well, after the definition, if you have a point P, this is O, this you find O P prime in such a way that O P times O P prime equals to R square, where R square is the radius of this circle. And what's very important is that if you have some kind of a circle which is passing through the center of inversion, let's say this circle, its image would be straight line perpendicular to diameter of this circle. We have proven that. So, basically, using the inversion, you can convert a circle into a line. Now, obviously, if you have this circle passing through this point, and this point has image somewhere, so the line will pass through this point, right? Or, alternatively, if you have this circle which is, let's say, tangential to this circle, and in another theorem, we have proven that any circle which is not passing through the center will be actually converted, transformed into a circle. Then the result will always be a circle tangential to, in this case, line. So, this circle goes into a line, this circle goes into a circle, but the fact that they are tangential, which means they have only one point in common. Now, if they have only one point in common, then these will have one point in common. You can't avoid that, right? So, that would be tangential ones. Alright, so that's basically the facts, short reminder of what we have learned before. And that's, and the inversion is something which we are going to use in these, well, now relatively simple problems. So, the first problem is, so we'll talk about, we'll talk about points and circles as given to us. And then we have to construct a circle which is tangential to these, or passing through the points. So, we have three elements which needs to be given. So, let's say, the first one is point, point, and, well, and point we have already discussed before when we were talking about points and lines. So, let's just talk about circle, point, point and the circle. So, we have to construct a circle which is passing through two points and is tangential to given circle. Let's say c. So, this is the circle we have to construct. This is the circle which is given and points are also given. Alright. Now, here is what I suggest to do. Let's take this particular point as a center of some kind of a big circle. And what happens if I will transform relative to this circle, this is my inversion circle and this is my center of inversion. So, what happens with this picture? Well, point A, it's a center of inversion, it doesn't participate in anything. Point B would go somewhere here, B prime. Our circle will go into some circle here, whatever the circle is. And what happens with this line? Well, this line in theory will be, since it goes through the center of inversion, it would be a line, a straight line, but it will still go through the point B prime and it will still be tangential to a circle c prime. So, that would be my line somewhere here. So, if I will transform this picture outside of this circle using inversion, point will go to point, circle to circle and line would be an image of this circle which we have to construct. So, what does it mean? It means that if I will transform a point in this circle and we'll be able to draw a tangential line to a circle, then I can transfer everything back and that will be my solution. So, what remains to actually to discuss? Well, number one is how to construct a point which is inverted image of given point and how to construct a circle which is inverted image of a circle. Well, that's simple, actually we were talking many times about this. So, we know that for instance AB times AB prime equals to R square where R is something which I know, that's the radius of this circle, that's my choice how to use, how to choose the R. So, we know AB, we know R and all we have to do is AB prime, right? So, AB divided by R is equal to R divided by AB prime and this is what we have to construct and this is a simple thing, you just have any angle, this is AB and this is R and this is R and then if you have parallel line here and here this would be AB prime from the similarity of these triangles, right? So, all you need to do is to have this segment AB then segment R on both sides actually because it's R and R here and here and then draw a line parallel to this line which is this. So, that's why B star, B prime. This segment is the one which we need. So, that's how you construct a image of the point. Now, how to construct image of a circle? Well, very simply, just go through the diameter, go with the line which connects my point A with the center. So, you have diameter, so you have two points. So, these two points can be separately transformed. This would be into this and this would be into that somewhere. So, you have two points and have a circle as a circle constructed on the diameter and the last but not least, how to construct a tangential line? Well, obviously, we have already discussed this many times but I will repeat. It's good exercise and again very simple one. Okay, so, you have a circle and you have a point and this is a tangential line which needs to be constructed. Well, you know that this is the right angle, right? The radius to the point of tangency makes the right angle with the tangential line. So, which means what you can do is build a circle on AO as a diameter because this is a locus of all points from which segment AO is viewed at angle 90 degree. So, you have two interception points and these are two solutions, obviously. So, this is one tangential line and this is another tangential line. So, that's basically it. So, we had two points in the circle and we have solved this problem using, let me repeat it again. First, you draw some kind of a big circle, you transform everything from inside of the circle outside, both points and the circle, draw a parallel, draw a tangential line from a point to a circle and then transform everything back. Well, not everything, just this line, you have to transform. How to transform a circle into a line? That's, I did not really say. So, if you have this, obviously this would be your point which should be transformed back here using the same technique which I was talking before about. So, you know AB prime in this case and you know the radius and you have to find AB. So, that's simple and then on AB as a diameter you just draw a circle and that's the circle which we need and that's the circle which would be the solution of our problem. Okay, now let's go to the next problem. So, we had point, point and circle. Now, let's consider point, circle and circle. It's actually as easy. So, you have a circle, you have to construct the circle which is going through the point A and is tangential to two circles, given circles. Let's say this is C, circle C and this is circle D, point C, point D. So, we need to construct this thing. I suggest to do exactly the same. Let's have a big circle around the whole picture with center A. Now, if A is a center then transform everything outside. So, what will happen? This will be a circle, circle C will go, this is D prime, circle C will go into a circle C prime. So, now what you have to do is, now this, since it goes through a center of inversion, it will be a mutual tangential line. So, we know everything about this except how to construct tangential line to two circles. Again, in some previous lectures we did discuss this problem. It's actually a very easy one. But after we construct this, we just transform everything back and that's a solution. So, let me just stop for five seconds how to construct a line which is tangent to two given circles. Well, let's consider you have this and this and you have to construct this line. Now, this is perpendicular, this is perpendicular parallel to each other. Right? So, let's do it this way. Let's reduce, now if this is lowercase r, this is capital r, let's reduce this circle by lowercase r. Basically, what you have to do is draw a parallel line and this will be a new radius. So, new radius would be r minus r, capital r minus lowercase r. Right? Because this is r, so this is r. So, from capital r, we subtract lowercase. Now, what about this line? Well, since it's parallel to this one, it's perpendicular to this one and obviously, it's a tangential line to this circle. Right? So, how the construction supposed to be working? So, this is an analysis. So, construction is you have these two circles, just reduce one of them by this radius, by the smaller one radius, then from the smaller center draw a tangential line and we know before how to do it. So, now, all we have to do is draw this perpendicular here, parallel to this one here and that's the tangential line for both of them. All right. That's all for points and circles given. Next lecture will be lines and circles and then all of them, like one point, one line and one circle given. So, that will be next lecture. That's it. As you see using the inversion, all these problems are really much simpler. All you have to do is to know how to transform from inside of some circle outside the whole picture, points and circles, whatever is given to you. Well, that's it for today. I suggest you to very thoroughly examine whatever the material, whatever the notes for this lecture are on unizord.com and then just put it aside and try to prove everything just by yourself. And it will be very helpful if you remind yourself all these simple problems like how to construct a tangent to a circle given the point and the circle, how to construct tangent to two circles and some other. For instance, how to divide an angle in two, how to divide a segment in two because all these are very elementary things and you just have to basically know it almost by heart. All right. That's it for today. Thank you very much and good luck.