 Hi, I'm Zor. Welcome to Unisor Education. So let me continue with my favorite subject, construction problems in geometry, especially those related to circles, tangents, angles, whatever. They are very, very useful and I very much hope that you have already solved all these problems yourself before listening to this lecture. If you didn't, just stop listening to me and try to solve all of them yourself. They are easy and it's just the process of solving these problems really what's needed, much more than just listening to my solutions to this. Because the purpose is for you to develop your creativity, your ability to put different ends together and come up with a solution, not to listen to whatever the radio solution I provide. Well, at the same time if you have already solved, then by all means please listen to whatever I'm saying and just compare, I mean, there are different solutions to the same problems without any questions asked. Now, so let me just continue. This is construction number four lecture, and it has 10 problems as usually, my standard. All right, let's start. Construct a circle that contains a given point and is tangent to a given straight line at a given point. Okay. So you have line and a point and another point. And you have to have a circle which is tangent to the line at that particular point and also going through this line, through this point. So you have the line and the point A on it and you have the point B. You have to build this circle which is tangential at point A to a line and going through B. Okay, this is one of the examples of using the locus, two locuses actually to locate one particular point because one quality which this particular point, the center of a circle which we're looking for, possesses is that if you draw a perpendicular, if you connect it to the point A, it will be perpendicular to the line. The radius to the point of tangency is perpendicular to the tangent itself which means that this center is supposed to lie on the perpendicular to a given line in a given point. So this perpendicular is something which we have to build first as the locus of all points from which we can use as a center of a tangential circle. At the same time, this same center is supposed to have another property. It's equidistant from A and B because these are both radiuses, right? Which means where is the locus of points equidistant from two given points? Well, it's a perpendicular bisector between them. So this is the sector, this is the segment and this particular line is a perpendicular bisector of the segment AB. So one locus and another locus. So this point is supposed to belong to both of them. So that's why it's on their intersection. So obviously, now when we have finished analysis of this problem, the construction itself is simple. Number one, we draw a perpendicular to our given line to the given point. Number two, we have this segment AB because B is also a given point and do the perpendicular bisecting. Now the intersection of these two locuses is the center of a circle which we are looking for. And the distance from A and from B is the radius. Knowing the center and the radius, we know the circle. Next, number two. Construct the circle tangential to both legs of the given angle if one point of tangency on one of the legs is given. So you have to draw a circle which is tangential to a given circle, not a given angle to both legs if one of the points of tangency is given. So you have an angle and you have this point. Now, let's think about this point, the center is every distance from two different legs of an angle. Now, why? Well, for all this reason, because it's tangent to this, so the radius to the point of tangency is perpendicular. So this is the radius, the distance from the point to, from a center to a leg. Now here, the same thing. From the radius to the tangent, it's perpendicular, so that's the distance. Now, so where is the locus of points equidistant from two different legs of an angle? Well, obviously it's angle bisector because any point here will be equidistant from both legs. Why? Because this is the hypotenuse. These are right angles. And since this is the bisector of the angle, these two angles are congruent to each other. So triangles are congruent, and that's why these casualties are congruent to each other. Okay, so this center is supposed to be on the bisector of this angle, which angle is given so we can draw a bisector. That's one locus. Another is, let's think about it, this point is located on, well, this point is supposed to be located on the perpendicular to this leg at this point. We have this point as a given, and this is a perpendicular, which means that this is another line, the perpendicular to this leg at this point where the center is supposed to lie. So we have two lines, the intersection of these two lines is a circle, a center of the circle we would like to construct. So we found the center, and we have the radius actually, so we have the whole circle. They are more easy as you see. Given two straight lines parallel to each other and point between them, lines and point between them. Construct the circle tangential to both lines and containing the given point. So we have to build a circle which is tangent to both lines and going through this point. Again, analysis. This center, since it's the center which is on equidistant from these two parallel lines, and again, why is it equidistant? Because this is, this radius to the point of tangency is perpendicular. This radius to the point of tangency is also perpendicular to tangent. Now, these are two parallel lines, which means this is basically a diameter. And no matter where this particular circle is located, this length is fixed and the center is in the point of this segment. So what we can do is we can say that this particular center is always in between these two lines on equal distance from it. So we have a mutual perpendicular to these parallel lines divided in half and draw a perpendicular bisector to this segment. So this is the line which is also parallel to these two and lies right in the middle between them. Everywhere on this line, we can put a center and a circle will be tangent to both parallel lines. Now, out of all these lines, we need only one, out of all these centers, we need the one whose circle actually going through this point. Now, how can we do that? Well, let's think about it. We basically know the radius right now, right? Since this is half of the distance between parallel lines. So we know this distance. So the center is supposed to be on this distance which we have determined on this distance from the given point. Now, where are all the points which are in the given distance from a given point? Obviously, this is a circle around this point. So wherever this circle of this radius, which is half of this distance between the parallel lines, wherever this circle intersects our midline between these two parallel lines, which is this and this two points, that's where the center of the circle which we are looking for is supposed to be located. So this is one of them. And this is another solution when it actually going through this line but on another side of a circle. So two solutions exist in this particular case. Analysis actually gave us one locus and another locus. One is a straight line, another is a circle around the given point of the radius which we have already determined. Intersection between these locuses give us the solution. Construct a tangent to a given circle such that it forms a given angle with a given straight line. Okay. So you have a circle, given circle. Now, we have to construct a tangent which has a given line, a given angle, this is an angle, okay, with a given line. So we have a circle, we have a line, and we have an angle. So we have to draw such a tangent which makes this particular angle we put to this one with the given line. And by the way, there may be more than one solution in this case, but let's talk about one particular solution. Okay. So we know that this particular angle is given, right? We also know, obviously, that this is perpendicular to the point of tangency, all right? Now, what can we say? Let's just choose any point on this line and using this angle, draw a line at this particular angle. It will not be tangent, obviously. But, however, what we can say is that the real tangent which we are looking for and this line which has exactly the same angle, these are parallel, right? Because these are equal congruent angles. So these are parallel, and this is transversal. If these angles are the same, lines are parallel. Now, what does it mean that the lines are parallel? It means that the perpendicular to one of them is perpendicular to another. So, since we have already built this line, we can always, from the center of the circle, draw a perpendicular to this constructed line. And wherever this perpendicular intersects the circle, is our tangency point for the tangent which we are looking for. So, as soon as we got this point, we draw a perpendicular in this point to the radius which exactly is the tangent which we are looking for. Well, now, why there are multiple solutions? Well, number one, this actually, this line, this radius which is perpendicular to this line, is intersecting circle in two points which means this is another tangent which also makes the same angle alpha with our given line and is also a tangent. On another side, we can have the line which makes an angle on another side. This would be alpha. And then we can draw a perpendicular to this and it also crosses our circle in two points which means we can again put one line and another line. Both are tangents and both are making angle alpha equal to this one with the given line. So, it looks like there are four solutions, at least. I don't think maybe there are more. It looks like only four. All right, so let's enter this. Let me make one quick point. In all these situations, I usually draw a circle or a line or a tangent or whatever else as already kind of ready for me to analyze. So, then I'm doing analysis and I'm saying, okay, this particular, let's say it's a center of a circle which we're looking for, is supposed to satisfy these properties. And then, when I have found that through the analysis that there are certain properties that this particular center is supposed to satisfy which means it's supposed to be on this local center that locals, then the real construction actually starts. Okay, I draw a locals, I draw another and then the crossing of these two lines or crossing of the circle in the line or whatever else I'm getting my answer. So, the whole problem solving actually is done in like two phases which I'm bunched together actually as one solution. First stage is analysis and then second stage is real construction. So, let me try maybe next problem to differentiate actually these two stages in more clarity, I would say. Okay, given a circle and a point outside it, construct a second from a point to a circle such that the cord formed by it was a congruent to a given segment. Okay, circle, point, second, and we want this particular segment to be congruent to some given segment. So, we have to draw a second from a given point to a given circle such as a cord which is being cut by a circle from a second is congruent to this particular segment A. So, how can we do this? Well, let's think about it. If we know the circle and we know AB because this is our A, obviously as you know, if you have a cord most likely you will need to put the radius perpendicular to the cord. And let's think about triangle KBO. What do you know about this triangle? Obviously you know the radius which is R and you know KB, which is one-half of A where A is our segment. So, you know the catatose and hypotenuse of this triangle. So, you know the triangle. You can construct it by hypotenuse and a catatose. This is one-half A and this is R. This gives you this particular segment, OK. The segment OK you built basically. You construct by constructing this particular triangle, right? Now, is it sufficient? Basically, yes. Because what you have to do really after that all you need is draw a circle concentric with the given one of a radius B. And now you see that MA is tangent to this small circle, right? So, this is analysis. I can find out the OK. I can then circle around O with the radius equal to B and I can draw a tangent to this smaller circle. All right, so this is analysis. Now, how the construction actually is doing? OK, here is the construction. Now, you don't know this line. We have to draw it, right? All we have is a circle, a point and a segment. So, what do I do first? First of all, using the radius, I know the radius of the circle since circle is given and half of this segment, which is half of A, I built a right triangle. Now, we all know how to do it, right? For instance, you can build the right angle. You cut half of A here. Then you have in the compass the R using this as a center. You basically cut the R and this is your B. OK, so you build this triangle so you know this segment. You take the compass, you take this segment, use it as a radius of a new circle here. Now, what's next? Next, I said we have to draw a tangent from a point to a circle. Now, you do remember how to do it. It's one of the previous tasks. You basically connect them together and divide it by half and use this as a circle. So, wherever it crosses, since this is the right angle, right angle is inscribed into this bigger circle and supported by half a circle, that's why it's right. And that's why we use OM as a diameter. We found the center of this diameter and drew a new circle and that's what makes the angle OKM the right angle. So, wherever this circle intersects this smaller circle, this is a point K. And Mk actually gives you the second we are looking for. That's the construction. So, first I did some kind of analysis, then using the analysis I construct the second which we are looking for. Construct a circle of a given radius tangential to a given straight line and containing a given point. A circle of a given radius tangential to a given straight line. OK, so you have a straight line. Now, let's consider that we have built this circle of a given radius and we also have a point. So, we have a line, we have a radius and we have a point. All right, so let's just think about where exactly this particular center of a circle which we are looking for is located. On one hand, since this distance is known, this is the radius, it means this center which we are looking for is supposed to be on the line parallel to the given line and on the distance equal to the radius, right? So, every point on this line can be a center of a circle of this radius which is tangential to the line. At the same time, this center is on a distance r from a given point which means it's supposed to be on a circle around that point of a radius, same radius r. So, wherever this line dotted line and this dotted circle cross each other, that's where our center is supposed to be located. So, this is an analysis. Now, the construction itself, all right, so this is the line, this is the point which is given and this is the radius. So, first I construct the perpendicular, I cut the radius, the perpendicular again or parallel to this line. And then, the same radius from here, I'm circling around my given point. So, these are two centers which can, two points which can serve as centers. So, from here, this distance is r and this distance is r which means that this is a tangential to the line and containing this point. Same thing with this, this distance is r and this distance is r. So, this is another circle which is also tangential to the line and going to the point given. That's it. Okay, given a circle and a straight line outside it, a circle and a straight line outside it, find a point on the line that the tangent from it to a circle has a given length. So, if you find this point, this is a tangent, it should have a given length. Well, obviously, you have to draw a radius to the point of the tangent, of the tangency and what you know is this is an r and this is the a. Okay, let's just wait until my phone stops ringing. So, this is the right triangle because this is a tangent. This is the radius to the point of tangency so they are perpendicular. The a, b, we know that's the given and the radius is given. So, the triangle is given and o, b is given, well, not given but it's basically constructable as well. So, you construct this triangle. Now, I'm going to a construction stage. You construct this triangle by two categories, r and a. This is a. This is hypotenuse. Now, having the hypotenuse, you basically draw a circle around o using this hypotenuse and the radius and wherever it crosses the line which is given that's where the point from which we should really draw a tangent to a circle is located. And from this point, the tangent, this one and this one, by the way, would have exactly the same lengths as well as this one and this one. So, we have two points and four different tangents which basically make up a solution to this problem. Instruct a triangle by an angle an altitude from the vertex of this angle. Instruct a triangle by the angle an altitude from this angle and another altitude from another. Now, what do we have here right now? So, we know the angle and we know two altitudes, vd and ce. We have to construct a triangle. By the way, I don't know, this is probably just my attempt to make some kind of a repetition of whatever we did before because this is just triangles there are no circles here. However, I found this problem and it might actually have a certain connection to circles. But anyway, let's just think about how to solve the problem. So, we know that there is an ec and vd two altitudes and we know the angle. Alright, so, basically, if you will take a look at the triangle e, b, c, it's the right triangle and we know two elements of this right triangle. The acute angle b and we know the characters e, c. So, in theory, we can always build triangle e, b, c knowing whatever we know. Angle and this altitude. Alright, so, let's just start building. So, for instance, we have built this particular triangle using this angle and the characters e, c. Now, what's next? Now, we have this altitude which means that... what can we say? So, we have to do this. Okay, we have the point c. We don't have point d. We have a distance between b and this line which means this line, if I will draw a circle, okay, here is the circle connection. Now, I understand why this is here. If you will draw a circle of the radius bd, center b, it will be tangential to ac, right? So, as far as our construction going on, so we can do is, from the b, as from the center, we built a circle of the radius equal to our second altitude. And, since cg is a tangent, now, and we already know how to do it, we built a tangent from c to this circle and that would be our point g. And that would be our point a. So, that's how we built it. First, we built this triangle by an angle and a catheter. Then, a circle around the vertex of the radius equal to the second altitude and then the tangent from this point c, which we have built in the very beginning as part of the ebc triangle, the tangent points g and continue to point a intersecting the continuation of ebc. So, that's how we get the ebc. I was wondering why we have triangles. Now, I understand why I put it there. Not just a repetition, okay? Given two points, construct a straight line such that perpendicular to this line from given equal to two given segments. Okay, let's say you're analyzing this problem. So, you have to find a line with these two distances given. So, you have this distance and you have this distance. So, the distance from a point to a line is along the perpendicular. So, these are two perpendiculars. They're parallel to each other, by the way. But in any case, it's also very useful here to have circles arranged because if I will have a circle from this is the center and this is the radius, it will be tangential to the line. Same as here, also tangential to the line. So, if we know these two radiuses basically these two circles, we can draw the circles and what do we have to do now? We have to do a tangent which is common to these two circles. So, we have to be able to draw a tangent to two circles. Okay, let's consider this problem separately. Now, let me forget about these circles and let's concentrate on this problem again. What if I will draw a line parallel to this line? Again, we're back to analysis. What we know right now is this is, if this is A and this is B, this is B minus A, right? So, what we know right now is that if I will draw a circle here of a radius B minus A, then my problem is much simpler. Now, I don't have to really draw a line tangential to two circles. I have to draw a line which is tangential to this smaller circle and originating in this point, at this point. Now, this is the problem which we have already solved many times, how to draw a tangent from a point to a circle. We know how to do this, right? You use this as a diameter, draw a circle around it and wherever it crosses, this is the right angle which means this is a tangent. Okay, so we know how to draw a tangent from a point to a circle. So, now we know how to approach this problem. First, you don't have this line yet. Now we're in the construction stage. Okay, so first, you calculate B minus A, the bigger minus smaller, and use this as a center. That's number one. Number two, so this is B minus A. Number two, you draw a tangent from this point to a circle. Okay? You know how to do that. What's now? Now, we can draw a perpendicular to the point of tangency. Now, this is B minus A. Then we cut A and we draw a parallel line here. So, since these are both perpendicular to the same line, these are parallel. This is also A. And this distance is A. This distance is B minus A plus A, which is also B. And that's what actually is needed. So, we draw a line which is on a certain distance A from the point A and B from the point B. And the way we do it is by drawing a circle, drawing a tangent and shifting the tangent down by certain lengths. Okay? And the last problem in this lecture construct a circle tangent to a given circle at a given point and containing a given point. So, there is a point on a circle. What we have to do is we have to draw another circle which is tangential to this circle and going through a point. Now, two circles are tangential to each other if they have only one common point. And we know that the perpendicular to this point where they touch each other is basically a tangent which is tangent to both of them. We solved that problem before. All right. So, what we do know is we know construct a tangent. Okay. So, basically having this point we know this line. All right. The center is supposed to be in continuation of the line which connects the center of a given circle and the point of tangency. At the same time the center is equidistant from these two points. Both are given points which means it's supposed to be in the perpendicular bisector of this segment. All right. So, one of the analysis the construction is so given a point on the circle and another point. So, what do we do? First, we draw this. Secondly, we do a perpendicular bisector to this and this is the center and this is the radius. That's it. That was my last problem Well, again, I hope it was educational. Don't forget that Unizor is the place where you can find these and many other very interesting problems. Solve problems yourself. That's very, very important before you listen to these lectures. And also parents and supervisors and the teachers who would like to control the educational process of their students are welcome at Unizor. It provides you the ability to enroll your students into particular programs check their score on exams which they should take and basically mark the particular lectures or programs whatever as completed or not completed depending on the results of the exam and try to make sure that your students are really going through the maximum possible score on every exam. It's very important. All right, good luck and thanks for your attention. See you next time.