 Hi, I'm Zor. Welcome to Unizor Education. I would like to continue talking about similarity. We went through a similarity of different geometrical objects, primarily consisting of straight lines like triangles, polygons, etc. Now we will go to circles of this extension. Well, if you remember my first lecture about similarity, I was actually talking about something like, well, intuitive understanding of what similarity actually is. And intuitively, circles are similar. I mean, all circles are similar to each other. They're all round, right? Well, let's just approach this mathematically. And to do it, I would suggest the following program. Number one, we will talk about what happens with a circle if you scale it. Well, obviously I will prove that if you scale a circle, you will get a circle of a different radius, but still a circle. Then question is, are all circles similar to each other? And the answer is positive, yes. I will basically build the scaling mechanism for proving of the similarity. So let's just do it step by step. And then a couple of side issues will be also addressed. So the first theorem is, scaling of a circle results in a circle. All right, so let's consider you have a circle and you scale it using this as a center. Question is whether you will get a circle as a result. Let's just consider factor equals two just for this particular example. Okay, first what I will do, I will scale the center and see what exactly would be the result. And let's say the result would be the point P, because the point Q should go along the same line, but twice as far from the center of scaling as Q used to be. And now I will do exactly the same with A. And I will get point one. Now let's consider this segment and this segment. Now, as it was proved before, any segment is transformed by scaling into another segment, which is parallel to the original one. Parallelism is very important in this particular case. And the second property is that the length of Bp equals factor times length of AQ. Well, the factor is two, Bp is twice as big as AQ. Now, we also know that the triangle XAQ is proportional to, is similar, and that's why every element of it is proportional to XBp. So XQ is elongated by the factor of F, and XA is elongated by the factor of F. So Bx relative to Ax equals Px relative to Qx and equals to factor F. So basically these two triangles are similar, and that's why this AQ is stretched by the factor of two in this particular case. And what's very important is that if I will take any other point A' which will be converted into point B', I can also say exactly the same with B'p equals F times A'Q. It was exactly the same consideration just considering this particular triangle. Now, all we have to do now is think about AQ or A'Q or any other distance from point on the circle to the center. It's a radius, so all these segments have equal lengths. They are all radiuses. This is a radius and this is a radius. Which means these two also are equal to each other and they are equal to F times radius, where radius is radius of original circle. So whenever I take any point on the original circle, I will get another point somewhere on a circle around image of a center of the radius which is factor F greater than the original radius. Which means all these points B and B' and any other lies on the circle. So from a circle, from the image of all the points on the circle, we will get points which are lying on the circle around the image of a center. But here is a very important consideration which you have to really think about. If I will take all the points in this circle, question is are images of these points after scaling, will they fill up completely the circle or I will have holes somewhere which are not actually images of any point here? Well, the answer is that the filling will be complete and there is a one-to-one correspondence between the points on this circle and points on that circle. And to prove this is true, what I will do is the following. Let's take any point on this particular circle and apply scaling with a factor equals to 1 over F. And obviously it will be shrink to some point here because actually it's exactly the same as before. It's just different factor, doesn't really matter. So from each point from here in the original factor I get some points from there but if I will use factor 1 over F, obviously from any point on this circle I will get a point here which actually will be the point which if scaled by the original factor F will go to this one. That's why I'm basically saying that for any point here there is a source on this circle which will be converted into this point after the original F. And how to find this point? Well, take this point and apply 1 over F scaling. So first you apply 1 over F, then at the point then if you apply the F it will return back to this and that's how we found that for each point in the image there is a point in the source. So it's a complete transformation of circle into a circle whenever we are scaling. It's actually extremely simple theorem and I spent like too many words probably to explain it but I want it to be relatively rigorous and especially I would like again to point out the necessity to prove that every point on an image will actually be an image of some point in the source. So we will completely fill out the circle by scaling this circle. Okay, enough said. Let's move on. Any two circles are similar to each other. How to prove that? Well, if I have two circles I can congruently put one inside another so they will share the center. Now this congruent transformation does not really change the similarity because similarity is defined as a combination of some congruent transformation like shift, rotation or reflection. And scaling. So let's forget about scaling for a second. I congruently transform one of my original circles into the bigger one so they have the same center. Now I will build explicitly the scaling to transform one into another. Well, obviously I will take the radius of this and the radius of this. This will be my center of scaling and the ratio between radiuses will be my factor. So obviously any point on this circle if I will multiply the lengths by this factor I will get the point on that circle. So using this particular transformation of this as a center and ratio between two radiuses as a factor using this I will convert any point on this circle into some point on that circle. And again, as in the previous case I would like to point out the completeness of this transformation in the sense that if I will take all the points from here and transform them using this scaling into points over there then all points of that circle will be filled up. How to prove it? Exactly the same way as before. Well, let's consider any point there and find a source from which I can obtain that point. So now for each point I have for each point in an image I can find the point on the source circle. So that's how we proved that any point on an image is actually the result of transformation of some point on the source. That's why we completely fill up the image. So now again I have applied first a congruent transformation. I shifted and just shifted we don't need to rotate anything. So I shifted the center and then I built explicitly the scaling. Now, next theorem is a bit more interesting. Question is, is there a scaling which can transform one circle into another without shifting? So let's say I have this circle and this circle. Is there a transformation which will transform this into this without shifting? Yes, with the shifting we can do it. Just take the center, make it here coincide, shift the whole and then just do the scaling. But what if you don't want to shift? Well, the answer is yes, there is such a transformation. And again I'm going to explicitly build this transformation. Well, but does it always exist? Actually not always because if you will take a different picture and this one, well, there is basically no such transformation in this case. So my point, my theorem actually is the following. If you have two circles and one is completely outside of another, then there are more than one actually ways to transform using scaling one into another. Let's talk about one particular way. Let's have a common tangent and connected with the central line. So the common tangent means that these are both perpendicular. So what I am saying is the following. I can use this point where the common tangent is intersecting the central line as a center and ratio between radiuses as a factor. And then using this type of scaling I can transform this into this. And that's what I'm going to prove. Okay, now, first of all, there are many different tangents, quite frankly. There is one tangent. There is another tangent. There is this type of a tangent. And there is this type of a tangent. These are all common tangents. So my first statement is these two tangents, which I can call one-sided because both circles are on one side of the tangent, they are intersecting the center line in the same point. So it doesn't matter whether you take this one or this one. But how can I prove that? Well, it's actually quite easy. Why? Because let me put letters p and q, m and n. And this is x, let's say. Now, this is the common tangent which means p, n is perpendicular to it and q, n is also perpendicular. So these are all parallel lines. These two radiuses are parallel. Which means that these angles correspondingly are equal to these angles which proves that triangle x, p, m is similar to x, q, n because two angles are correspondingly congruent to each other. Which means that the length of x, m relates to the length of x, n as smaller radius to the bigger radius, right? Now, exactly the same thing can be said about these two triangles. Let's call it p prime and q prime. mp prime is perpendicular to this line and q prime is perpendicular. So they are parallel. So these angles are correspondingly equal to each other as well as these angles. Doesn't really matter whether they are equal to those or not. They are, but it doesn't matter right now. So they are equal to themselves, right? So, again, these two triangles x, m, p prime and x, m, q prime are similar to each other because two angles are correspondingly congruent to each other. Which means, again, the length of the point of intersection it must be the same because if I will mark it x prime, for instance, because x prime to m relates to x prime to n exactly the same as, again, as two radiuses. So since ratio between x, m to x, m should be exactly the same as x prime m to x prime m, from this it very easily follows that x and x prime should coincide with each other because these ratios are the same. Okay, so forget about this common. We just incidentally proved that two common tangents are intersecting the center line in the same point. So doesn't really matter which point we take. What we do have to prove is that if I use this as a center of scaling and ratio between radiuses as the factor of my circle with the center m would be converted into circle with the center m. Alright, how to prove? Let's take any point A here and let's say B is its image and I'm not actually saying I've put it on the circle but I'm not saying it's on the circle. It's some point B. And let's now talk about mA and mB. Now, since B is an image of A using this scaling then every element which is xA, for instance, to xB or xm to xm or am to bm they are all in the same ratio which is ratio between two radiuses. But am is a smaller radius which implies that if I will multiply by the ratio between bigger radius and smaller radius I will get the point B on the distance mB equals to am times, let's say, R capital R over lowercase r ratio between the radiuses which is a factor. So, regardless of the position of point A since am is equal to the radius mB would be equals to the radius with the capital R, obviously, right? Because am is equal to lowercase r. So, all the points which are images of any point on this circle lie on that circle. And that's how I prove that the image of a circle is basically a circle. Well, almost proved because, again, I have to prove that every point on the larger circle can be obtained from some point on a smaller circle. Now, how to do it but exactly the same as I did before. Take any point from here and apply a reverse ratio lowercase r over capital R and you will get some point here which, if scale using the original factor would give this one. Now, is it always possible to find a center of scaling in this case the way I just did? No, because if these two circles are of equal radius then, obviously, this common tangent would be parallel to the center line and they will never intersect. So this will not work. However, this would work. So let me now concentrate on other common tangents which I can call two-sided common tangents because both circles are on both sides. So let's forget about these points and forget about this common tangent. So this is true for circles lying outside of each other and of different radius. But if you have the same radius the second method works well. And, obviously, it works with the different radius. So this second method is more general because it's applicable to circles of both equal and non-equal radiuses. Okay, let me just wipe this out and it's actually very, very similar proof. So you have one circle and you have another circle and you have not very beautiful but still and the center line. This is perpendicular and this is perpendicular. Now, once more if this is the point x this is m, this is m, this is p, this is q obviously these two triangles mxp and nxq are similar to each other because these angles are vertical and these angles are congruent to each other as this line is parallel to this because it's two perpendicular to the same tangent. So we have two parallel and transversal so these are alternate inside, that's interior alternate interior angles so they are equal to each other. So these two triangles are similar because they have two angles correspondingly equal to each other which means that the ratio between radiuses is exactly the same as ratio between mx and xn and now we'll do exactly the same thing let's take any point here and let's draw the line through x well actually in this case it's this one and once more consider these triangles not a very good picture actually but so in this particular case what do we know about these triangles amx and b and x well what we know is that they have ax proportional to bx as as what? as mx proportional to ax since b is an image of a relative to the center and whatever the factor is between the radiuses and m and m is image of m and these angles are vertical we have the similarity of a amx and bmx axm and xn are proportional to radiuses that's the way how we built it in the first place and these guys are proportional by our construction and that's why this is proportional to this in exactly the same ratio which means bm is equal to am multiplied by the same factor which is ratio between the radiuses so it's exactly the same proof so from any point here we get the point which is on the same distance from m and all I have to prove basically this circle is filled up completely by images of points from this circle and again the way how I do it I just use the reverse factor instead of radius the smaller radius I will use smaller to the bigger and from this point any point on this circle I can build the point on this circle okay so this second method is as I was saying applicable even in case circles are of equal size alright if there anything else I'm missing or actually no yes this theorem completes my just general discussion about circles and basically I think all the theoretical material which I wanted to present as a lecture is complete on similarity so from now on I will start basically doing something which is part of Unisword.com which is solving problems and obviously you understand that everything I was talking before about definition of similarity and different similar objects and basic theorems is just a preamble to solving problems which is the most important part of this website this is the language this is the syntax if you wish the semantics is where the problems are that's the purpose and that's what I'm going to do in the next lecture now meanwhile I do recommend to review this material on Unisword.com and as far as the problems are concerned they are in the next lectures always try to do it yourself first before you listen to the lecture I'm trying to address this to the parents as well because the parents can control the educational process of their children or group teachers for instance they can control the educational process of every member of the group on an individual basis and that's what I would like to encourage adults use this website for that's it for today thank you very much