 Hi, I'm Zor. Welcome to Unizor Education. I would like to continue talking about circles, about small theorems, mini theorems, which are plenty. All of them serve only one purpose, to train your brain to do certain creative things, invent certain additional constructions which you should really make to prove your point, or to reduce maybe your more complicated task to some, you know, more manageable, smaller tasks. Which is a great feature, by the way, which you can use anywhere in your practical life. So, brain exercises to continue. Okay, this is a series number three among mini theorems about circles. All right, prove that the shortest distance between two circles, lying outside of each other, is the segment between them lying on the line connecting their centers. Okay, so you have two circles outside of each other. These are centers, m and n. So, what I am trying to prove is that the shortest distance between these two circles is this particular segment, which is obtained by connecting the centers, and these are intersections of this centerline with both circles. So, first of all, before actually addressing this particular problem I would like to say a couple of words about something which I was just using, the terminology, distance between circles. Well, it's not, you know, easy to talk about distance between different objects if we don't define it, right? Have we defined it before? Well, inadvertently, we probably did. However, I think it makes sense to repeat this generalized approach to distance. If you have two different geometrical figures, each one of them basically is a set of points. Now, you can always choose a point on or inside or wherever of one object and another point from another set of points. So, there is a distance between pair of points, one point being a member, an element of one set and another point being an element of another set. Now, the distance between two different geometrical objects, which are two different sets of points, is the minimum which you can find if you can of this particular distance. So, if there is such a thing as a pair of points, let's say this and this, such that the distance between these two points is less than or equal than any other distance of any other pair of points, then this distance is called the distance between objects. So, in this particular case, to prove that the shortest distance, C is between A and B, all I have to do is prove that if I will take any other two points, one belonging to one circle and another belonging to another circle, then the distance will be greater. And quite frankly, it's very easy to prove. Let's consider you have two points, one here, C, and another here, D. And let's consider this distance from C to D. Now, I would like to prove that C is greater than A, B. How? Simple. Again, connect with the center and you know that M, C plus C, D plus D, N, which is an angular connection between M and N, is greater than M, A plus AB plus B, N. Because this is a straight line and this is angular connection. Now, if you consider these two, M, C is a radius, M, A is a radius, M, C and M, A. So, they can go out from this inequality. Same thing, D, N and B, N. D, N and B, N. They are congruent to each other. Both reduces of a small circle. And what's left? C, D, greater than AB. C, D, greater than AB. And that's what's necessary to do. So, what's the most important part? Well, you connect the endpoints to a center and use an extension to a triangle inequality which says that any angular line is connecting two points is longer than the straight line which connects these two points. Simple, isn't it? Next. Given two circles with centers at points P and Q intersecting each other, okay, now we have intersecting circles. P and Q intersecting at B and D prime. Okay. Draw two seconds that contain point B at the intersection but do not contain D prime. Okay, so two seconds. One parallel to PQ. So, we have a PQ, a center line between two. Now, one second is parallel. It goes through B parallel to the center line. This is one second. Okay, A and C. Okay, so ABC is a straight line. It goes through this intersection of two circles and it's parallel to PQ. AC is parallel to PQ. Okay. Then let's have any other second which is not parallel. Let's say this one. Proof that the segment AC is longer than DE. Okay, okay. So, what should we do now? Well, let's draw perpendicular. Whenever you have a cord, most likely you have to draw a perpendicular from the center to this cord. So, let this be letter K and this would be letter L. Put it on the top so it doesn't really confuse us. Okay. And as you know, every radius which is perpendicular to a cord divides it in half. So, basically, right from here I can tell you that the KL is equal to half AC. Why? Because this KB is half of this piece and the L is half of this piece BC. So, that's why if you add these two to get KL you will have half of AC. Now, if I will do exactly the same thing with this particular cord, DBE, I'd like to use the same letter, same using on the web, so you will not get confused. Right. So, for the same reason, MN measures half of the DBE. Why? Because MB is half of the DBE and BN is half of the DBE. Now, instead of comparing AC and DBE I can compare their caps, KL and MN. Right? All right, whose numbers? Now. Now, what's interesting is that this half of the AC, which is KL because AC is parallel to PQ, KL is equal to PQ since KL QP is rectangle. Right? So, PQ and KL are the same. So, instead of comparing KL to MN I can compare PQ and MN. And basically this is already down because if you will consider what MN QP is, let me just draw it again. MN perpendicular, perpendicular QP. All right? Now, MN and M, MP and MQ are parallel to each other because they're both perpendicular to the same DBE. So, these are parallel. So, this is the trapezoid. Just moved on the side, right? So, this is the trapezoid. Now, MP and MQ are parallel and MN is a mutual perpendicular. Now, you know that between two parallel lines perpendicular is the shortest distance, the shortest segment which connects point from here to here. That's why MN is less than PQ. PQ is not perpendicular to these lines. I mean, maybe if I will put it this way it would be easier. These lines are parallel. This is MQ. This is PM. All right? I just turned it sideways. Two parallel lines and obviously the mutual perpendicular is the shortest distance. That's why we have this inequality between PQ and MN. So, PQ is greater than MN. KL, therefore, is greater than MN and KL is half of the AC which means AC is also greater than GE which is twice as big as MN. Okay? That's it. Given the point P outside of a circle with a center O and two tangents from each two circles. Okay? Whenever we are talking about tangents we are talking about perpendicular radiuses to the point of touching. Tangency, point of tangency. AB and this is O. So PA and PB are two tangents. I draw OA and OB because I definitely know it will be necessary to do because whenever you talk about tangents you talk about radiuses which are perpendicular to them. So, prove that segments AP and BP are congruent. AP and BP. I don't know. Folks quite obvious. Obviously you have to consider these two right triangles with a common hypotenuse. Now these are right because these are right angles and these are two radiuses. So you have hypotenuse, common and radius, two tangency are equal to each other, congruent to each other. In the right triangle that's sufficient for congruence of the other tangency. That's it. That's the simple thing. Number four, given two circles with only one point of intersection that is tangential to each other. So we have two circles which are tangential to each other. So as you see not only the line can be tangent to a circle. A circle can be tangent to a circle and the definition is exactly the same. The line is called tangent if it has only one common point, one point of intersection and the circle is also called tangent to another circle if they have only one point of intersection. So if they are far from each other there are no intersections. If they are touching each other, tangent to each other there is one point of intersection and if they are even closer you will have two points of intersection obviously. All right. Two circles, one point of intersection called P. I'm sure we will have to connect the centers. Okay. How many different lines exist which are tangent to both circles? Well obviously this one, this is one, this is another. Obviously in between there is another line. So this line is also tangent to both as well as this as well as this. So there are three different lines which are tangent, each one of them is tangent to both circles. This one in between the circles and these two lines are touching the circles on one side. All right. So we can call this one interior and these two exterior tangents but doesn't really matter. Prove that interior tangent divides in half segments of the exterior. So we have to prove that this interior tangent divides segments AB and CG in half. So we have to prove that these two are equal and these two are equal. Now quite frankly it's very easy. I have just proved in the previous task, the previous problem that if you have a point and two tangent to the same circle then these segments have the same lengths from the congruency of these triangles. So let's consider point M. Well obviously M A and M P are two tangents which are originated at point M and that's why A M is equal to M P. Similarly, since M P is also a tangent to this particular circle M P is equal to M B. So that's why A M and M B are the same. Similarly from the point M we originate one tangent and another tangent. So these two segments are the same and we originate these two tangents to this circle and that's why P N is equal to N D. And that's why C N and N D also are congruence to each other. So each segment the points of tangency is divided by half by the interior tangent. Next, given two circles with centers M and N outside of each other and two lines each tangential to both circles. Okay, A and B, C and D. Okay, so we have two circles we have two tangents C D. When you have two circles most likely you will have to connect them the centers. Don't see. Okay, prove that segments A B and C D are congruent. A B and C D are congruent. Okay, let's do it this way. If this is not a circle but a point let's consider this is a point like this one. Then we know that these segments are the same, right? So if you consider this point P we know that A P and C P are the same because it's two different tangents originated from the same point P. A P is A P is equal to C P. At the same time let's consider this circle and the same point P. Obviously B P is equal to D P. Now what A B is? A B is A P minus B P and C G is equal to C P minus D P. But since A P and C P are the same and B P and D P are also the same then the differences will be the same. That's why A B and C G are the same. By the way, that's not exactly the proof which I have put on the web. So this one just came to my mind right now. It seems to be easier than the way how I proved it on the web. But both proofs are quite rigorous so it's interesting actually to compare this one to the one which I have on the web. Alright? Easy. As you see, all our theorems are really easy if you know how to approach them and how to know how to approach them. Well, the more you solve, the more you prove the better you will be to basically come up with whatever the idea is necessary to prove this particular theorem. As you see, for instance, whenever I draw a tangent I immediately draw a radius to a point of tangency. Why? Because all the theorems and all the problems are about this particular perpendicularity of the radius to the point of tangency. If you have two circles, most likely you have to connect their centers. Maybe not, but it's always something which you have already done before. That's why it's very important to go through as many problems like this as possible because it will inculcate in your mind the approach you can take. If the same thing is working in 10 different cases probably it will work in 11 as well. Alright? Given two circles, tangential to each other at point K and exterior common tangent that touches at A and B. Okay, so one circle and tangential another circle. And we have this point K. And then we have an exterior tangent. These are perpendicular radius to the points of tangency. A, K, B, A, B. Probes that A, K, B is the right angle. A, K, B is right angle. This is not obvious, by the way. I mean, if all other theorems were more or less obvious the fact that this is the right angle is not that obvious, mind you. Now, how can we solve it? Well, something which immediately comes to my mind is if you remember we had a theorem that if you have a tangent and a chord which is originated from a point of tangency then this angle is half of this angle. Remember that, right? It's one of these previous theorems which I have proven. So what does it mean? Well, it means that, yeah, I think it will work. Again, it's not exactly the way how I put it on the web but it looks like it might actually work. So let's put this angle, let's call it alpha and this angle, let's call it beta. So alpha is one half of angle A, M, K. Right? A, M, K. It's a central angle and it is supported by the same arc as this chord which is originated at the point of tangency. Similarly, beta is equal to one half of B, M, K. If you add them together you have alpha plus beta is equal to one half of A, M, K and B, M, K. Now, but these are supplemental angles, right? Because these lines are parallel so it's like interior one-sided with these two parallels and M, N as a transversal. So their sum is 180 degrees. So we have 90 degrees. So alpha plus beta is 90 but now A, K, B is a triangle and triangle has 180 degrees as the sum of all triangles. So what's left for A, K, B? Well, 180 minus alpha minus beta which is 90. So A, K, B equals 90 degrees. It's not the way how I put it on the web. It's a different proof here. And again, this one just came to my mind because I realized immediately that we were actually talking about this theorem that an angle between a chord and the tangent where the chord has an end point is actually measured half of the corresponding center of angle and that's immediately going into this particular theorem. Well, it's interesting how different proofs are coming to mind when you are doing it two or three different times the same thing. So first I was doing it for the web and I just came up with this list of problems and solutions. And then as I'm explaining, some other idea comes to my mind and that's what makes the whole thing actually a very creative process. All right. Okay. Given a circle and a point M outside. Circle and point M outside. From M, draw two tangents. M A and M B. M A and M B. All right. On a smaller arc between A and B choose any point C and draw a tangent. Point C and tangent. Any point C on a smaller arc which is closer to M. It intersects at D and E. Other tangents. So this tangent intersects these two tangents at points D and E. Prove that perimeter of triangle DEN. DEN. Perimeter of this triangle is independent of the position C on the arc. So no matter which point C we will take perimeter of this triangle DEN would be the same. Yeah, well, it's obvious. From D you have two different tangents D A and D C which are congruent to each other. So these two segments are equal. Similarly from E you have this segment equal to this segment. So perimeter of D M E is equal to D M E perimeter of triangle. Let's put it this way. Perimeter of triangle D M E is equal to M D plus D C plus C E and plus E M. But D C is equal to D A. So let's substitute D A and C E is equal to E B D E. We substitute it D C for D A D C D A and C E B E. Well, obviously M D plus D A is M A and B E plus E M is B M. And this is a sum of these two things completely independent of the position of the C. Yeah, that's like this. And then you can have to draw any additional lines or anything. It's just, you know, straight consideration based on the fact that two tangents from the same point outside of a circle are equal to each other. Lengths, I mean. Okay, the last theorem is given two congruent circles lying outside of each other with centers M and M. So congruent circles have exactly the same radiuses M and M and the same radiuses. A second parallel to M and M intersects at points A, B, C and D. So this line and the parallel to this line A, B, C and D prove that A, C and M is a parallelogram. A, C, and M is a parallelogram. Okay. Let's do it. First of all, as usually if you have a chord you probably have to draw a perpendicular to this chord which is its perpendicular bisector. Okay, how do we call these points? Do we call it anyway? K and L. Okay, K and L. Now, let's think about it. K, L and M is definitely a rectangle Y because K, L is parallel to M and M. K, M and L and M are perpendicular to the same line which means they are also parallel to each other and since they are perpendicular by construction this is rectangle. So K, L, M, M K, L, M, M is a rectangle. Now what's next? For obvious reason triangles A, K, M and C, L, M are congruent right triangles. Why? Because the casualties K, M is congruent to calculus L, M because these are two opposite sides of the rectangle, right? And A, M and C, M are radiuses of congruent circles so they are equal in length as well. So they have a hypotenuse and the calculus congruent to each other. So triangles are congruent which means this is equal to this A, K and C, L A, K and C, L are congruent to each other. Therefore A, C which is equal to K, L minus C, L but plus A, K so it's K, L minus C, L and plus A, K now C, L and K, A are as I have just proven equal in lengths so we have the A, C is equal to K, L so the distance from here to here is the same as from here to here but this is a rectangle which means it's equal to M, N so A, C and M, N are equal in lengths and parallel to each other and if you remember if you have a quadrangle with two opposite sides parallel and equal in lengths to each other this is parallelogram which is exactly what is necessary to prove. Well that's it, that's it for this particular series of mini theorems about circles everything you can find on Unisor.com and as I told before there are certain proofs which are different over there than I have just used during my lecture but it's just because you know something came to my mind and it seemed to be quite rigorous and interesting and educational if you wish it's always interesting to find something on the fly that's what creativity is all about alright I do encourage all the parents to take a look at the website and use it to basically control the educational process of their children because you can enroll children into certain course or part of the course