 Hi, I'm Zor. Welcome to Unizor Education. I continue construction problems in the geometry of circles, my favorite subject. I do presume that number one, you have already solved whether yourself or with the help of the previous lectures, all other previous construction problems which have been presented on the website. Because primarily the reason is some later problems are using the results of the previous ones, as ever St. Kelsen mess is just built one upon another. So if I will refer for instance to some older problems which we have already solved and you solved yourself, then I don't really want to go into the details about how to solve that problem. I'll just refer to the previous one. All right. So without further ado, let me just go this is number five set of construction problems for circles. All right. One by one. Hopefully do it fast. Construct a circle tangential to two given parallel lines and to another given circle between them. All right. So you have two parallel lines and a circle in between them. What you have to do, you have to construct a circle which is tangential to these parallel lines and to this particular circle also given to you. Now, what does it mean that the line is tangent to a circle? Well, the first is obviously the perpendicular to the point of tangency is this radius. Radius into the point is perpendicular to the line. Now, same thing here. Now, these two lines are parallel to each other which means this is one and the same line perpendicular to both parallel lines. So this is a diameter because it goes from the center. It's two radiuses, right? So it goes from the center and since the lines are parallel, these two radiuses are making one diameter. So this is one thing, mutually perpendicular to these lines. Another is if you connect the centers of the new circle which you have to construct and the one which is given, the center line is intersecting both circles exactly in one point of tangency, which is common to both circles. So this is some kind of analysis. Now, question is how can we build this type of thing? All right, what I think is probably the easiest way is instead of constructing this circle, let's construct a bigger circle concentric to this one. It will look like this. Now, this radius is bigger than the one which we need by the radius of the small circle which is given to us. Now, why is it easier to construct a bigger circle? Well, because number one, we know it actually contains the center, a point rather than being tangential to a circle. It's easier to construct a circle which contains a given point rather than construct a circle which is tangential to a given circle. It's just obvious that it should be easier this way. Number two, if the smaller circle is tangent to this line, then if we increase the radius of a small circle by a known segment, which is the radius, obviously it will be tangential to another line parallel to this one, which is also on the same distance r from the one which we had before, and same thing here. This is also r. So, now after this analysis has been basically completed, here is a solution which I am offering. Number one, you shift the top line by r. r is a known value because that's the radius of the circle. You shift it up, shift down the parallel line, the bottom parallel line also by r. Now this point, it used to be before. So, these are parallel lines which are given. The dotted parallel lines are the shifted up and down. The point, this is the center, and we used to have this circle and we still have it. But now, what we have to do is we have to construct a circle which is tangent to these dotted lines and passing through this center. Now, this is actually quite easy because number one, we know that this particular center of the circle which we are looking for is supposed to be in the middle between these parallel lines. So, we can just take a perpendicular and bisect it. So, this is the line where our center is supposed to be located. So, that's the next construction. And on another hand, since we know the distance between these lines, original lines, between these two lines, this is actually the radius of a circle which we are looking for, which means if we will take from here with a radius equal to the radius of this circle which is r, plus the radius of a circle which we are looking for, which we know as well. And using the sum of these two segments, we just draw a circle around it with this as a center. So, this point is on the distance of the radius which we are looking for plus the small radius. And that's actually exactly what we need. So, this is an intersection between this circle and the middle line between these two parallel lines. So, this is how we can construct this bigger circle. Well, actually, we don't really need to draw it because since we have a center, we can construct immediately the smaller one and it will be tangent to a circle and both original parallel lines. So, what's important here is to basically expand the circle which we are looking for by these ranges and the whole problem is converted into a simpler one to have a circle which is tangent to two parallel lines and is passing through a particular point easier than the one before. And by the way, this is also true for every problem of that kind. You basically try to simplify the problem by reducing it to something which you have already solved before or which looks really easier. Now, this problem looks easier if instead of having a circle which is tangent to a given circle, you have to have a circle which is just passing through a given point. Point is simpler than the circle, if you wish. This point is simpler to a circle. So, you simplify the problem and that's the key to everything. Okay, that's it. Next, construct a circle of a given radius that contains a given point that lies inside a circle or on a circle or outside a circle. All right, so we have a circle which is given and we have a point which can be either outside or on a circle or in the circle. Let's talk about outside first. So, construct a circle of a given radius so we do have a radius as well. And using this radius, we should really build a circle which is tangential to this circle and passing through this point. Something like this. And this radius is, so we have to construct this circle. What's given is this circle and the point and the radius. Well, let's put it this way. It's actually quite easy because this point is supposed to be on the distance r from a given point, which means if you want to construct and this is a. First of all, where are all these points, centers of a circle which we're looking for which are on the distance r from a? So this is the locus of them. So you draw a circle of a radius r around the given point. So somewhere on that circle, this center is located, okay? At the same time, this center is, if this is lowercase r, this center is on the distance lowercase r plus capital r from this center. Again, using this combined length of its own radius plus the given radius, I can draw another circle. These are all the points which are on this distance from the center. Now, obviously our point which is the center is supposed to satisfy both criteria. It's supposed to be on the distance r from the a and on the distance r plus r from o. So this point of intersection between these two circles is the one which we're looking for. And by the way, this point is good as well because the circle here and the circle here, both will do the trick of being tangent to our circle and also passing through this point. Many of these problems actually very much look alike and there are only like small distinction. You have a point which your circle is supposed to be going through or you have its radius or you have maybe another line which is supposed to be tangent and all these conditions are just played in different combinations. That's basically all. Right, now, slightly different is if this point which we are given is on the circle, not outside. So we have a radius, we have a circle and a point but it's not outside as before but on the circle. Well, in this particular case, look at this drawing. Now, obviously, since a circle which we're looking for is supposed to be tangent to this circle, it means it's supposed to have only one common point. At the same time, point A is supposed to be contained in this circle but it's already contained in this circle because it's on it, which means that the point A is one and only point of tangency between these two circles. So all you need to do just connect over to A center to the given point, expand it by this radius and here you get the center. So that's easy too. And the last part of this problem, what is the point which is given is inside the circle. So if this is point A and again you have some radius and you have to build a circle. Well, let's draw a circle and then we'll see how it goes. Something like this. This is the tangent point. So this is the center of original circle and this is the center of a circle which we're looking for. This is tangency. Now since this is tangency and this is R and this is lowercase R. So all C is equal to R minus R. The radius is lowercase R and the smaller radius is capital R which is given which means that the distance from O to C is R minus R. So the locus of all points which lie on this distance from the O is this circle. So this is our first step in construction. At the same time, the locus of all points which are on the distance R, this is capital R from the given point A is this circle. And again we have exactly the same situation as in the case of point outside, except outside it was a sum of two radiuses. In this case it's the difference between two radiuses. So the intersection of these two circles will give you two different points which can be used as centers. Okay, so all three variations of this problem we have solved. Construct the circle of a given radius, tangential to a given straight line and tangential to a given circle. So again, all the different combinations. So you have the radius and you have circle and a line. Now, how do you find this point? Well, it's basically very much like before. You do two locuses. Locus number one is locus of all the points which are on the distance R plus R from the center of a given line. So if you have this and you have this, this is your R. So using the center and using the R plus R, its own radius plus the given, you draw a circle of this radius. That's number one. Number two, since this distance is supposed to be equal to R because from the center to the tangent, you know the radius to the point of tangency is perpendicular, so this is the distance. Now, what is the locus of all the points on the distance R from this line? Obviously it's the parallel line which lies along the perpendicular on the distance R. So the intersection between this line and there is another intersection down there with a circle, with this dotted circle, these are intersections, these are centers of the circle which we are looking for. Again, two locuses. One is the distance from the center another is distance from the line. Both are given, so the crossing of these two gives you the center. Construct a circle of a given radius that intersecting two legs of a given angle forms two quarts of two given lengths. Circle of a given radius. So there is a radius, there is an angle, there is a circle which we have to build in such a way that this chord and this chord have given lengths. This is AB and this is CD. So the lengths of these two quarts are given and the radius is given. So how can we build this circle? All right, let's do some analysis here. Obviously, if I will connect the center with the midpoint of a chord, it will be a perpendicular, as you know. Whenever you see the chord, probably it's the good thing to connect the midpoint of this chord with the center and it will be a perpendicular. Now, what do we know about these right triangles? This one and this one. Obviously, if this is midpoint M and this is M, obviously MB is half of AB, which is known. Similarly, MB is half of CD, which is known too. So we know the cavity and we know hypotenoses. So we basically know these triangles because these triangles are right triangles and to construct the right triangle, you need just two elements. In this case, the catatose and the hypotenuse. So by building this triangle, which is our first step, we determine OM. Similarly, we can determine OM by building triangle OMD using again the catatose, which is half of this, and the hypotenuse, which is the radius. So OM is also determined, constructed. So these are our first two steps. Now, after we know MO and MO to construct is basically easy because MO is distance from this leg of an angle. So on this distance MO, we draw a parallel line, which is along the perpendicular is equal to MO. And from this leg, the distance from the center is OM, which we also have determined. So we draw another parallel line on the distance of OM. And intersection of these two parallel lines gives the center. Again, two locuses and their intersection. What was the most important part of this? To determine the distance from the center to one side of an angle and to another side of an angle. And to determine this, we have to just construct these right triangles, which obviously I'm not talking about how. I'm sure you know it from the previous lectures, triangle, right triangle can be constructed by a catatose and hypotenuse, that's easy. Okay, next. Construct a circle tangential to a given circle at a given point and tangential to a given straight line. Okay, so we had a similar thing actually. So what's given are circle with a point and the line. The previous problem, which was kind of a similar, at least the drawing was very much similar. We didn't have this point, we had the radius of this circle, which we have to do. Now, this point, it's different, but obviously the first thing which you obviously can think about is that this center is supposed to be on this line, right? That's kind of obvious. All right, now we don't know the radius, but we do know that the center lies on this particular line, on this particular, actually it's a part of the diameter. All right, fine, so how can we find the center now? Let's think about it. Now, this is obviously perpendicular to, this is the radius, very easily. If we will draw the line which is tangent in this particular point of tangency, as you know, this line is tangent for both, this and this circle. Now, we can draw this line because it's tangent to this particular circle in a given point, so there is no question about where to get this line. But now, look at this particular circle. This circle is touching both sides of an angle, which means it's supposed to lie on the bisector of this angle for obvious reasons. So now our problem can be solved in the following way. Let me do the construction itself. That's what's given, right? First, we draw the line, which is this one. That's where the center is supposed to be if this circle is tangent to this circle. So this is at the point A. So this is the locus of all the centers of all the circles which can be tangential to this circle at this point. Next, we draw a tangent, this one, and we know how to do it. It's basically perpendicular to this line at point A. And then we have a bisector, and this is the center of a circle which we are looking for, and this is the radius of this. Okay? By the way, as a side note, I'd like to mention that, well, you know, I was always talking about simplification of problems. Now, what we had done here, instead of this circle being tangent to this circle, I draw this tangent which is kind of common tangent to both. And now I'm saying, okay, this circle is actually tangent to the line. Line is simpler than a circle. And as you see, it was easier to draw a circle which is tangent to the line rather than to this particular sector because now we can use this bisector, angle bisector approach. With a circle, we don't really know that we can't really do this. So again, line is simpler than circle. Point is simpler than circle. Point is simpler than line. So whenever you can replace one problem with another, it would make your job easier. All right. Construct this circle tangential to given circle and to a given straight line at a given point. Okay, now situation is very much like before, but the point is not on the circle, the point is here. That's the point where it's supposed to to be the point of tangency. So it means obviously that the center is obviously on the perpendicular since this is supposed to be a tangent to this circle. The radius into the point of tangency is perpendicular. So since we know the line, the tangent and we know the point where the tangency actually occurs, the perpendicular to this line to this tangent in this point would be the locus of all the different centers which we're looking for. Okay, now that's one thing. Now, next thing is to basically to use this circle. Okay, and now what I'm going to do is exactly what they just said. I'm going to simplify the problem. Now, let's expand this circle by this radius r. What happens? Then the new circle will be like this and now instead of being a tangent to a circle, the new circle will go through the center of a given circle, right? And obviously it will be tangent to this line which is parallel to this given line and on a distance r. We just expanded the blue arc basically, this circle which we're looking for by the radius r which is given and now our problem becomes slightly easier. And what's easier about it, the distance between these two points is supposed to be the same. This is the radius of a big circle. Now we know this point because we know this point, so this is the mutual perpendicular and we know this point. So where is this one which is equidistant from these two, connect these two together and put a perpendicular bisector of this segment, right? And that would give us another locus which is intersecting with the one which we already had and that gives you the point, the center. Now again, I'm not actually going into the details how many solutions it has depending on how big the radius is, et cetera. And in all these cases, quite frankly should actually be discussed in more details. But I think they're all kind of obvious and they don't really need this additional construction, et cetera. The most important thing in this problem is just to understand what's the approach to get to the center. And then when this approach is once made, then the variations in different cases are really easy. That's why I don't really want to discuss them in details. The idea was, again, one locus which is perpendicular to this tangent at this point, another is blow up this circle in such a way that you have two points where this center is supposed to be equidistant. You found these two points by putting another line parallel to this one on the distance r and then from these two points you just draw a perpendicular bisector and then you get another locus. Now the last one in this lecture construct a circle tangential to three different circles of equal radiuses lying outside of each other. And consider two cases. All three are outside and all three are inside. Okay. So you have three different circles of the same radius lying outside of each other. What you have to do, you have to build a circle which is tangential to all of them. I'm not sure. I don't think I need this. I can jump ahead of myself. So we have these three circles and the one which is tangential to all of them. So this line obviously is intersecting at the point of tangency. This line and this line as well. Now let's not forget that all of these guys have the same radius. That's given. These are all radiuses which we don't know but they are radiuses which means they're equal as well. So what this particular problem therefore implies we have to find a point which is equidistant from three different points. Now we know basically how to create a point which is equidistant from three different points. It's basically like a circumscribe, a circle around the triangle, same thing. So you connect these two guys and put perpendicular bisector, because perpendicular bisector is a locus of points equidistant from these and this is a locus of points equidistant from these. So therefore their intersection is the point which is equidistant from all of them, all three of them. And that's exactly what's needed. Basically that's it. And the radius is obviously obtained by connecting to any one of those centers and getting this point. Now what if I would like to have another case when it's not inside, when it's not outside, when these three given circles of equal radius are not outside of this circle but inside? Well, I think it's very similar. Let me just draw it back. It's easier to just draw it from scratch. So I have three different points so I have three different circles and I would like to have a circle which is around it. So it's tangential to all of them but it's outside. Now again, let's connect these two guys to the point of tangency. Well, this seems to be a little bit more complicated, isn't it? However, the situation is pretty much the same. Why? Because, again, this point of tangency lies on the center line. So this is one line. O A D C D E. So O A D is one line. It's not angular because again, the point of tangency of two different circles, even on the inside tangency, is basically going through both centers. Same thing here and same thing here, which implies that the distance O A is equal to O D minus A D which is the radius and it's equal in all three cases. And the radius O D is also equal to all three cases. So in this case, the distance from O to A to B and to C is equal to R minus R in all three cases. Now we don't know R but we do know that it's the same R in all three cases, which means again, the point O is equidistant from all these three centers. And to construct it, you just do exactly the same thing. A couple of perpendicular bisectors to segments which connect the centers give you the center. Now once you have a center, again, connect the center to all three and you get all these three points and these and each one of them is a radius so you have the radius. So the construction in both cases is actually the same. You build the point O which is equidistant from these and both solutions are actually implying that this particular point is the same. In this case and in this case, point is the same. It's just different radius. That's it. Find the point, connect it and you have two solutions, one and another. Concentrate to each other because the center is exactly the same. It's exactly the same point which is equidistant from three given points. Well, that's, I think that's it, right? That's the last one. All right, now this is not the last lecture about construction. Even construction with circles, I still have some more and I will continue a little bit more complicated problems maybe. Don't forget all the problems which you see, you must spend some time to basically try to solve it yourself because it's just the process of solving which the whole thing is all about. It's about solving problems. You will not need anything from whatever I'm talking about right now in your practical life.