 Hi, I'm Zor. Welcome to your new Zor education. I would like to continue talking about circles, but in this case it's not about theorems. It's about certain other types of problems which I would like actually to present. It's construction problems, it's problems about locuses, this type of problems. They are slightly different, they are easy. I didn't even put solutions on the Internet, just the conditions of these problems and solutions I'm going to present right now. All right, so let me just start with all these two problems. A couple of problems will deal with one particular concept which I would like to actually define. You know that if you have a circle, this line which has only one common point of intersection with a circle is called tangent. Sometimes we are not talking about an entire line, but only a particular segment of this line from certain point outside to the point where it actually touches the circle. So the point of tangency from a given point outside to the point of tangency, sometimes I will call this particular segment a tangent segment or tangential segment. It's just definition if I will use during one of the problems. And I did mention it before, I'll just repeat, if you have two different circles which have only one common point of intersection, then these two circles are called tangential to each other. So this is a tangent circle to this and this is a tangent circle to that. Just terminology, I will use them every once in a while. Find the locus of points outside of a given circle such that their tangential segments to this circle are congruent to a given segment. Okay, so you have a circle and you have certain segment. Now, you are looking for points outside such that their tangential segment will be exactly congruent to this one. This point will also be of this type and this type and this one. So all these points belong to this particular locus. So all the points, what exactly they represent? What's the set of points all having the same property that their tangential segment to this given circle is congruent to a given segment? Well, let's just think about it. If you have a tangential segment, this is the radius to the point of tangency. What is this? Well, obviously this is the right triangle because the radius is always perpendicular to tangent. In this right triangle, this is the radius which is a constant and this is a segment which is given. So it's also constant. So all these right triangles have this property of having fixed cacti which means their hypogenesis is also fixed. So this particular distance is supposed to be constant because all these triangles are congruent to each other by one tangent and another calculus, by two calculus. And that means that their hypogenesis are the same. So what we can say is all the points which have this property are equidistant from the center of our circle and the distance is exactly the hypogenesis of a triangle which has one calculus equal to the radius of a circle and another calculus equal to our segment. Now, we have to prove it like both ways. One is that all these points, since this length is fixed, are aligned obviously on a circle with the same center and the radius equal to this hypogenesis. So all points lie on the circle. Now the question is the opposite. Are all the points on the circle satisfy this condition? Which means that I will take any point and draw a tangent to our circle. Will this tangent have the same length? Well, the answer is yes. Because again, this is the one calculus which is equal to a radius. This is a hypotenuse which is the same as in this case and in this case, et cetera, which means this triangle is also congruent to all other triangles which I have already drawn. Which means that any point, if I take it on a circle, will satisfy exactly the same criteria. So all points which are lying on this circle with the radius being the hypogenesis of this triangle are the locus of points from which the tangential segment is equal to our segment. And again, don't forget we have proved this going both ways. If this is the point which satisfies our criteria, it lies on a circle, on a bigger circle. And if it lies on a bigger circle, it satisfies our criteria. And that actually proves that these and only these points are those which satisfy our criteria, which means it's a locus. It's very important actually to prove these both ways that certain set of points which satisfy certain criteria are of a certain geometrical shape. And then all the points of that geometrical shape really have this particular characteristic which we're looking for. Then and only then we can say that this is a locus of points characterized by this particular property. Number two, find the locus of points outside of a given circle such that their two tangential segments to this circle form an angle congruent to a given one. So from any point outside, we draw two tangential segments. These are perpendicular, obviously. And this angle is supposed to be given. Obviously, as you know, these two triangles are congruent to each other by common hypotenuse and magnitude which are radiuses, which means these angles are congruent to each other and equal to half of our given angle. So considering this, if this is a given angle and this is a given radius of a circle, that necessitates that wherever I take this point, which satisfies our criteria, this distance must be the same. So wherever else, if I take these two tangential segments, then again, it's the same radius. These are all the same. And since the angle is supposed to be the same, a pattern is supposed to be the same. So it looks like all the points which satisfy our criteria are supposed to be on that particular circle around the same center and the hypogenuse being a hypogenuse of a triangle with one characteristic equal to radius and opposite angle, acute angle, being equal to half of our angle. All right, so we basically did research. Now the proof itself that this is the locus, which means, number one, if a point has this property, then it's supposed to be on a circle. And if the point is on a circle, it has this property. All right, so we basically did it. Considering the point has this property of having this particular angle between two tangent segments, the angle equal to given. It obviously means that it's supposed to be on this distance from the center. Now how to build this distance? Well, very easily. You have the radius, and you have an angle, right? The opposite angle, or this one, which is 90 degrees minus this one. It's the right triangle. So you build this triangle, and the hypogenuse is the distance which you're looking for. So since my statement is that all the points on the circle of this radius is the locus, so basically if the point is such as the angle is equal to a given, then obviously distance is equal to this distance. And on the opposite side, if I take any point on a circle which basically has the property of being on this distance from the center, then obviously if I will draw a triangle, now this angle will go so equal to this one, which is half of our given angle, which means that the total angle will be exactly equal. So again, all the points of this property are on the circle, and any point on the circle has the property. So it's a locus. The locus of points from which a given circle is viewed, if you wish it's like a view, view from a point. You view at certain angle. You view this circle at certain angle from this point. So the locus of all the points, our circle is viewed at certain angle, is a circle around our circle of a certain radius which can be built. Number three, find the locus of centers of circles of a given radius tangential to a given line. OK. So given a line and given a radius, now question is, where are the centers of all the circles of these ranges which are tangential to this line? This, and this, and this, et cetera. Well, it looks like it's a line parallel to our line on this particular distance, right? It's kind of obvious. Well, again, let's just prove. So my analysis shows that the locus of points which have the property of being centers of the circles of the same radius tangent to our line is the line parallel to a given line and on a certain distance of it, which is equal to this particular radius. All right. So indeed, if a point is a center of such a circle, then this is a perpendicular. Since it's a tangent, this is a perpendicular. And that's why the point is on this particular distance from the line. Otherwise, if I take any point on the line, now since this is exactly the same distance, if I will draw a circle using this distance, it will touch the line since the radius is perpendicular to the line at the very end. So the answer to this particular problem is the locus is the line parallel to a given line on the distance, which is equal to the radius. And by the way, there are two different solutions because the line can be on opposite side of the given line because these are also centers of circles. This line is also the locus of centers of circles of the same radius, also tangent to our line, but on the other side of it. So basically, the full answer is two lines on both sides of our given line on this distance. Now, instead of a line, we have a circle and also a radius of another circle, which is tangent to this one. So this is another circle, which is tangent to our and has a given radius. So we need the locus of all centers of the circles of this radius, which are tangent to our given circle. Now, since if this is the radius big r and this is small r, then the distance between centers is r plus r at fixed length. So it looks like all these centers are supposed to be on the same r plus r distance from the center. So it looks like the locus is a concentric circle on the radius equals to the given radius given radius of the big circle plus given radius from a circle, which needs to be tangent. Now, again, we can very, very easily prove both things. Number one, that any point, which is a center of a circle, which has this radius and tangent to our circle, is located on the distance r plus r from m. That's how it is. Conversely, if you take any point on that circle, then considering this distance is r plus r, then if this is a center, you can always draw a circle and it will be tangent to our circle. So this big circle, the dotted circle, is locus. Now, is it complete? Well, now, the answer is, in some cases, you might have another type of tangency, if you wish. You can have an inside circle like this. It also has the same radius, and this is capital R. But now, this distance is r minus r. So this is r plus r, this is r minus r. So this is, therefore, very easily proven another locus. So outside and inside, in outside case, this circle has a radius r plus r. Inside is r minus r. Now, you can answer a little bit more sophisticated question. What if r is greater than r? Let me try to draw it again, and you will see how it looks. So I'm trying to find all the different solutions to this problem. So let's say you have this is r. And what if r, lowercase r, is longer? Well, obviously, the solution outside of a circle is exactly the same. There is no difference, so it will be something like this. And obviously, the centers will be on the concentric circle with radius equal to the sum of big r and small r. But for the inside tangent, the situation is a little bit different. So for a smaller r, we get something like this, right? But for a bigger r, for a lowercase r bigger than the capital, it will be like this. So if this is r, this is r. So from this point, we go back. It will be something like this. And this is the center, so it will be like that, right? And now, the locus will still be a concentric circle, but this cannot be actually the radius because lowercase r is greater than capital r. However, if you would take it absoluterio, which is equal in this case r minus r, see this is lowercase r and this is uppercase r. So this basically is r minus r. That's what this particular segment is. And this will be the center. And then, again, you move around using this as a radius and you will get another locus of points, which are centers of circles of this radius. And they all will be like this, outside of this. Well, it's not like smaller circle will be inside the original one. It's original will be inside of a bigger circle. So it's still a circle inside. But this radius in this case should be given as an absolute value. So the answer is one circle of this radius will be the locus. And another circle is absolute value of this radius. If this is greater, then that's what we were considering the first time. If small r is greater than big, then the absolute value will help to get to the real point. What's interesting here, we should not be confused by the drawing which we have made. Because first we draw a circle and then another circle outside. We immediately came to a conclusion that it would be a sum of two reduces as a locus. But you see, we have been dependent on this particular drawing. We have drawn two circles tangent to each other, outside of each other. But we just completely forgot. Well, we didn't forget, but we could have forgotten the fact that this particular touching, the tangency between two circles can be not only exterior, but also interior. So let's have to get all the different cases. Otherwise, we will miss certain solutions to our problems. OK. Oh, by the way, I think it's very appropriate right now. I do have a note here, and that's why I just remember, to prove that if you have two circles tangent to each other, then the point of tangency actually is supposed to be on the line which connects the circles. Now, why? Well, you remember that all points outside of a circle are further than the radius. So if you take any other point here, let's say this is a point of touching of these two circles. The problem is, if it's not one and the same point, this will be longer than this, right? So you would like to shorten the distance between these two circles down to the minimum. Minimum is 0, where they touch, right? So if it's not a straight line, then it will be greater. And so basically these points will not be one and the same point. So that's the way how to prove it. There is a way to do it a little bit more rigorously, but this is the way to prove this particular fact, that the line which connects the centers should really intersect the both circles in the same point where they touch each other. Otherwise, we will just be in contradiction with triangle inequality. A segment of a given length moves parallel to itself, such as one of the end points moves along the given circle. So you have a segment which moves parallel to itself, and one end is always on this circle. Question is, what would be the trajectory of the other point? Well, you guessed it, it would be exactly the same radius of a circle, but the whole circle will be shifted by the length of this particular segment. Now, how can we prove it? Well, how to prove that certain things is on the circle? Well, points which are aligned in the same circle are equidistant from a center, right? So let's start from the center of this circle and have the same parallel segment from a center. Now, I'm basically making a statement that this will be a center, from a center you shift to a center of a new circle of the same radius, and that will be the lowest point. Why? Well, because this is parallelogram, we were talking about the segment which is moving parallel to itself, which means the parallelism is preserved and the length is preserved. And that's the sufficient condition for a quadrangle to be parallelogram. So that's why these two are equal to each other. This is the radius of the original circle. And this now, therefore, since this, no matter where this particular segment is located, let's say it's this one. It's still a parallelogram because this is parallel and equal to this one. So this one should be parallel and equal to this one. So again, the radius is always the same and it's always parallel. So it will be the locus would be the set of points which are equidistant from this point, which I have obtained by shifting parallel to the same, parallel to this segment, shifting the center of the circle. So the whole proof, actually, is based on the quality of the parallelogram that if two opposite sides are parallel and equal in lengths congruent to each other, then the quadrangle is a parallelogram and the other two sides would be congruent to each other as well. So this is the radius, so this is the radius. A segment of a given length moves such that one of its end point moves along one way over right triangle, while another end point moves along the other leg. So you have a segment of a fixed length, one end on one side of the right angle, and another end of this segment on another side, something like this. So what would be the midpoints of these segments as the segment is moving from this position to this position, always having both end points on two sides of this right angle? What would be the trajectory of the middle of this particular segment? I would like to remind you, Assyrian, if you have right triangle, and this is the midpoint of a hypotenuse, then the distance from this midpoint to all three vertices would be the same. Now, how can that be proven? Well, first of all, we have already proved it many times before. But in theory, if you want, I can repeat something like this. Well, if you take this midpoint of this hypogenuse, so these two segments are of equal lengths, and draw a parallel to a cataclysm. Now, obviously, you remember that the line, which is dividing one side of the triangle in half, parallel to the base, would divide the other side in half. So this is parallel to this, and this is the right angle. So this is the right angle. And now, what you know about this is, you don't know about this yet. You know about this triangle, and this triangle, that they're both right triangles, and they have one cataclysm common. They share it. And other pair of cataclysm are equal in lengths. That's why triangles, this one, and this one, are congruent. And that's why hypogenuse are equal. The same thing with this particular side, obviously, because this is equal to this, and this is equal to this. So all three are the same. So basically, what you're saying is that the center of the hypogenuse is a center of circumscribing circle, because the distance to three vertices is the same. Now, you can use it here, because in all these cases, hypogenuse remains exactly the same, because we're saying that this particular segment is moving such that one endpoint on one side of the right angle and another point on another side. So as this segment is moving, it means that the hypogenuse is always the same. All these triangles have the same hypogenuse. But now, the midpoint, as we know, has a distance to this vertex of this angle exactly the same as half of hypogenuse. Hypogenuse is fixed, so its hat is fixed, so this distance is always the same, which means that no matter how we position this particular segment, its midpoint would be on a fixed distance equal to the hat of this segment from the vertex of the right angle, which means that quarter of a circle would be the trajectory of the midpoint. Let me just draw it again a little bit cleaner, and you will see that this is true. So let me draw. So any line, whatever you... This is the midpoint. Then, how should I draw it? Something like this, that's the problem. These are different positions of our segment, this one or this one. These are different positions of a segment which is divided in half by a midpoint. The midpoint would actually be moving along this particular circle because this distance is always the same and equal to half of a hypogenuse. Okay, fine. Find the locus of midpoints of all chords of a given length in a given circle. That's easy. So midpoint of all chords of a given length. Well, if the length is given, obviously all these triangles will be congruent to each other. Why? Because the length is given, which means half of the length is also given. It's exactly the same as this one. And since this is the radius, it looks like all these triangles are right triangles and they are congruent by hypogenuse and the catatris, which means that the other catatris, which is this one, is also fixed. So all these midpoints of all these chords are lying on a concentric circle, the radius of which can be obtained by constructing this triangle by antiparticles, which is equal to the radius of our original circle, and half of the chord, the length of which is also given. And again, it should be proven both ways. I just shortened my lecture because it's obvious that every point which has this characteristic of being midpoint of a chord is on the circle. And then if you have a circle, then obviously by choosing any point and drawing a tangential line, at this point you will get the chord of the same length. Construct a circle of a given radius, tangential to a given circle, and a given line. So you have a radius, you have a given circle and given line, so you have to have a circle of this radius, you have to find, you have to construct this circle if this circle is given and this line is given and the radius is given, you have to construct this circle. Well, let's just think about the distance from this point to this, let's call this lowercase r, and this is uppercase r. So the distance from this to this is r plus r, which means this point b, which is the center of a circle which we are looking for, is supposed to be in the distance r plus r from the center of a given circle, which means the locus of all the points which are lying on this distance is the circle, right? Now, on the other hand, the distance between this center and the line should be, sorry, I'm using lowercase r, should be r lowercase r, which means it should lie on a parallel line, on a distance lowercase r from a given one, on this side or on this side. So we have two potential parallel lines, which are on the same distance r from a given line on both sides, where this center is supposed to be located and it's supposed to be located on a circle of the radius r plus r from the center a. So whatever the crossing of these two lines and the circle are, any intersection would fit. This is one, this is another, because this radius is exactly the same as this one, r and r. And if our r is large enough, then this particular circle might even cross it in two different places. For both lines, let me just draw another picture, and you will see that there are multiple intersections and all of them will fit the bill, basically. So if this is our initial circle, this is a circle bigger than that. Now this is a line and this is another parallel line and another parallel line. So what are our intersections? One, two, three and four, four different solutions. In this point we can do it this way. In this point we can actually, this is the center, right? So it would be like this. In this point and in this point, I don't know how to draw it. But anyway, whatever the intersection is, it fits the bill. Okay, construct the circle of a given radius tangential to two given circles. Okay, one circle and another circle. So we have these two circles on the left and on the right, and we have radius. Well, actually I don't really have to do it on the same line. Let me just draw it this way. Okay, so radius of this circle is supposed to do this. Let me make it more truthful. Something like this. These two circles are given. So obviously this point is supposed to be on this circle, which is concentric to one of the given circles, and the radius are one plus four ksr. Now this is supposed to be also beyond this circle, which is of the radius r2 plus lowercase r from this center. So we have two different lobeses. Locals of all points on a distance r1 plus lowercase r from this point, which is a circle, and on the radius r2 plus lowercase r of a circle from this center. And wherever these two dotted circles cross, that's actually our point where the center of a new circle can be located. Now, don't forget that there are really many solutions to this problem, and it's all related to certain interrelationship, because the tangential circle can be either exterior or interior. So we have to really think about a drawing which looks differently, which looks like this, something like this. So this circle is the one which we are looking for. If this is the radius, these are two circles which are given to us. So this is just one of the solutions which can be obtained in case two original circles cross each other. If they have intersections, then we can do it this way. Or if you wish, if this radius is really very, very large, we can have even different solutions. We can have one given circle, another given circle, and if our circle, which we're looking for, has a very large radius, then it can be something like this. So two original circles will be interior to the one which we're looking for. But in any of those cases, this radius, this point, can always be found as an intersection of concentric circles of the radius either r1 plus r1 minus r2 plus r and r1 minus r. So any of those cases, and by the way, in these cases, you can put the absolute layer as well. So all these circles of all these radiuses are actually supposed to be considered because depending on their inter-relationship, you can have many different solutions to the same problem. So there is definitely more than one way to skin the cat. OK, I have the last problem. A circle of a given radius is rolling along a given straight line. OK, so you have a straight line. You have a rate of, let's do it more accurately. Let's start with the circle. And this is the line. So circle is rolling like a bicycle wheel on the road. This circle is rolling. Now, there is a point on the circle. And the question is, draw approximately a trajectory of that fixed point as the circle actually is rolling. OK, this is not really the problem which we can solve right now in all the details. I'll just give you approximation. And again, the problem is asking about approximate picture, not the precise picture. Because the curve is really quite complex one. If you will think about this, it will move down, but it will roll forward. So basically, it will be like this. Until finally, it will touch when the length of this will be equal to the length of this, it will touch the straight line. And then it will start rising again. And backwards, it goes the same way. So that would be an approximate curve, approximate trajectory of the point, which is fixed on the circle rolling along the straight line. And this particular curve has a name, say, Chloe. So it has an interesting properties. It was researched by certain mathematicians because it's complex. It's really complex curve. It's not a circle or a semicircle, not at all. The algebraic equation which actually describes this might be a little bit more complex than we are ready to do right now. But at least you know what the cycloid actually is. So this is a trajectory of a point which is moving on a circle rolling along the straight line. OK, that's all I wanted to talk about today. Thanks very much for listening and participating. I would like to point you again to the Unizord.com as the source of many interesting problems. And parents are very much encouraged to take charge in the educational process of their children by rolling them into certain programs, categories, whatever we have on the website. And let them take exams. And basically, if exams are fine, just mark this particular program as completed. If not, ask your students to do it again until they will perfect their skills. Thanks very much. That's it for today.