College Algebra: Word Problems Involving Circles

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Okay, I thought we'd close our circle of circle questions by looking at a couple of interesting examples where I think they are a little interesting. Some of them are kind of amazing. I'll write them up here on the white board. Let's take a look at them.
The first one is, to find all points, x, y, such that first of all, x = y. Okay, so they're equal and the same thing. Second of all, that they are all four units from the point (1,3). So I'm looking for all the points that are four units from the point (1,3), but for which x and y are the same. Well, if you think about it, what are all the points that are four units from the point (1,3)? Well that's a circle of points. So in fact we're trying to find points on a circle when the circle is centered at (1,3) and has a radius of 4.
So, first of all, let's just find the circle that's centered at (1,3), a radius of 4. Well, let's see. We know what the formula is for a circle. It's (x - h)^2 + (y - k)^2 = r^2.[ ]So what do we know? We know that the center of the circle must be at (1,3). So I'll insert that in for the h and the k - (x - 1)^2 + (y - 3)^2 =--and the radius, well, it's suppose to be four units from this point. So the radius is going to be 4, but 4^2 is 16. Okay, well there's the formula for all the points that actually satisfy that second part, which are four units from the point (1,3).
Okay, but now the points we're looking for have an extra property - do you see it? The extra property is that x and y are equal. So, what I have to now do is say - Okay, if x = y, then what does that mean? That means that I can sort of replace the y by x. So, let's do that. Well, then I would see (x - 1)^2 + (x - 3)^2 = 16. Right? I'm just saying if x and y are the same I can replace this y by an x. Now I have this. Well now I have just an equation with just x's in it. So I can solve this.
Now how would you solve this? Well, what I'm going to do is I'm just going to foil everything out here and see what this thing looks like and then try to combine and factor and it's a quadratic. I'm going to now square this out. x^2--I'm going to do it in one step, but I'll talk through it. Now, you have to visualize now an (x - 1) right here. So, I see a -x, another -x - that's a -2x, -1 times -1 is +1. Now I add a similar thing here - x^2 - 6x + 9 and that equals 16. Well, that's just a big old mess. Let's combine things a little bit. I see I have one x^2 here, another x^2 here. So that's 2x^2. How many x's do I have? I have -2x and I have a -6x here, so that's a total of -8x. What do I have here? I've got a 1 constant and a 9 constant, so that's a 10 constant. If I bring the 16 constant over to the other side that's a -16. But I have a 10, so that's going to be -6 = 0.
Notice I can divide everything through by 2. I have a common factor of 2 everywhere. If I do that, I would see just x^2 - 4x - 3 = 0. Let's see if this can be factored or not. So, this is x, x - the negative sign tells me they're opposite signs. So I've got a minus and a plus. I need something that's product is 3, but combined to subtract to give 4--and I don't know about this. Because if I put in a 3 here and 1 here that's only going to give me a -3 and the same thing here. So, I think actually this one is not going to be factorable. Sorry. So, I guess I have to use the quadratic formula. If I use the quadratic formula, see how that would play out.
With the quadratic formula on this thing I'd see that x = -B. So, there's the quadratic formula or course. -B--so that'd be negative -4, which is 4 plus or minus the square root of B squared--which would be -4^2 or 16 minus 4 times AC. Well, A times C would be -3. So that would make this negative sign a plus, times 3, all over 2A. So, what I would see here is 4 plus or minus the square root of--and I have 16 + 12--which would be 28 over 2. So, that's the x value and since x and y are equal, I see that in fact this is also the y value, and so what I conclude is the following. And there's two answers here, of course. So, what are the two answers? So, for which points do we have these things that satisfy the question? Well, the answer would be 4 plus the square root of 28 over 2, itself, 28 the 2, and the same thing with the minus in between there. 4, minus the square root of 28, over 2, 4 minus the square root of 28 over 2.
So these are the two points that satisfy those things. By the way, what does this look like graphically? I mean, should there be two answers? Maybe you think there should only be one answer. Well, if you think about it - I think there really should be two answers. Because if you have a circle, let's see, its center... Lets see if I can draw this somewhat accurately. I'll do my best. It's centered at--what was the point again? A (1,3) - so (1,3) and its radius is 4. So, it looks sort of like this, and then we want to see where in fact x and y are equal. Well, x and y are equal actually along this line--sort of goes like this. These are all the points where x and y are equal. Look, (1,1), (2,2), (3,3), (4,4). These are all the points where x and y are equal. You see (0,0), (-1,-1). And you can see that in fact this circle crosses the line at two points and those are the two answers that we actually found.
So, there we answer that question - sort of an interesting question. Now let me close with one last question. This question is find the equation of the circle having smallest radius that contains the two points (1,4) and (-3,2). Let me draw a picture of this first. So I have (1,4), (1,4). So there's (1,4) and then I have (-3,2), (-3,2). So there's (-3,2). If you think about it there are actually a lot of circles - a lot of circles that pass through those points. In fact there are infinitely many circles that pass through those points. Let me show you some. Here's a real big one. I can even get a bigger one. But I can get smaller ones too. I can get one that sort of looks like this. Here's a smaller circle, but I want the smallest circle. If you think about it, to get the smallest circle - that means that these two points should be as far apart as possible on the circle. Which would mean that these two points should be opposite points on the circle. So in fact the circle should look something like this. Well, not very ovally. These two points should be opposite points.
If they are opposite points then the center of the circle should be right in the midpoint of those two things. So to find the center, I find the midpoint. So midpoint would equal the center. And what would that be? Well I average the x's. So I take 1 plus -3 and I get -2 and divide by 2. Then I average the y's, which is 4 plus 2, which is 6 divided by 2. So see that the center is (-1,3), which looks pretty good, by the way. Look, (-1,3) looks actually very good. What would the radius be? How would I find the radius? Well, the radius would just be in fact the distance between one of these points and the midpoint. So, I'll use distance formula. So radius would equal what? Well, it would be the square root. So this point now--let me put this in for you. This is point (-1,3), that's that point right there.
So, the distance between these, I take -1 and subtract -3, that's -1 minus -3. So that's -1 plus 3, which would be 2, and then I have to square that number. Then I add to that the difference in the y. So, 3 minus 2, which is just 1 squared. So, that's the square root of 4 plus 1, which is the square root of 5. The radius of the circle is the square root of 5. We were asked to find the equation of the circle. I know its center, I know its radius, so the equation I can write down as what? x minus and then the x point, which is a -1. So minus a minus is a plus 1 squared, plus y minus the y point, which is 3 and this equals the radius squared. So, what's the square root of 5^2? It's just 5.
So in fact this is the answer. This is the smallest circle that contains these two points. That's pretty cool. If you wanted to find the smallest circle that contains two points - just make them be opposite points. If you think about it that's as small as you're going to possibly get. They are as far apart as possible. Therefore the midpoint, that must be the center. You can compute the radius by looking at the distance between one of the midpoints and one of these endpoints you found. Anyway, enjoy yourselves with circles. They are the most beautiful, symmetric things around.
Relations and Functions
Circles
Solving Word Problems Involving Circles Page [2 of 2]