WEBVTT
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Now let's look at circles.
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A circle is a set of points
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that are a fixed distance from the center (h, k).
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Here, (x, y) represents any point along that circle.
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The letter r represents the radius of the circle,
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which is the distance between the center and any point on the circle.
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Using the distance formula, we can get an equation.
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Let's look. The distance is going to be equal to
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x minus h squared plus y minus k squared.
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Of course in this particular example,
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that distance represents the radius, so we'll write: equals r.
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Now let's just look at this right here.
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What happens if we square both sides?
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We're going to get x minus h squared
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plus y minus k squared equals r squared.
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This is very important.
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It is the standard form of the equation of a circle.
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You need to remember this formula and what everything stands for.
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Remember (h, k) is the center,
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(x, y) represents any point on the circle, and r is the radius.
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Now if we were to multiply all this out,
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and bring all the terms to one side and leave zero on the other,
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we're going to get another form of the equation of a circle.
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It's going to look like this.
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This is called the general form of the equation of a circle.
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You need to remember these two forms.
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When they ask for a particular one, pay attention to which form you put it in.
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Now let's look at an example where we find the equation of a circle given information.
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Here, we're given the radius and the center,
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and we want to find both forms of the equation of the circle.
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Let's start with the standard form.
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Recall the standard form looks like this:
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x minus h squared plus y minus k squared equals r squared.
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Here, h and k are (-2, 1), and the radius is 3.
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Plugging that in, we get x minus a negative 2, which is x plus 2,
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and we get y minus 1 squared, and we get 3 squared, which is 9.
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Here is the standard form of the equation of the circle.
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Now to get the general form, we're going to multiply all this out
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and then get zero on one side.
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Let's square x plus 2.
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We're going to get x squared plus 4x plus 4,
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and we're going to square y minus 1.
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We get y squared minus 2y plus 1 equals 9.
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Now we're going to collect all the terms on the lefthand side.
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We're going to write it in the form: x squared first,
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then the y squared term,
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then the x term, then the y term,
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and then we're going to get all the constants together.
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Notice on the left, you have 5,
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and when you subtract 9, you're going to get a negative 4.
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Here is the general form of the equation of the circle.
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Let's look at another example.
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Let's look at this example where we're given the equation of the circle,
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and we want to find the center and the radius of the circle.
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Graph the circle, and find the intercepts if there are any.
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Well, notice first of all you might be tempted to think that this is in standard form,
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but remember in standard form,
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the coefficient of the x squared and the y squared is a 1: the x minus h.
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What I want to do first is to divide everything by 3.
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Now it's in standard form,
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so now I can say that the center is, and remember it's (h, k),
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so h must be zero, so the center, (h, k),
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is 0 and y minus k, so k is 5.
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The radius is the square root of 4, which is 2.
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Now with that information, we can easily graph the circle.
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Here's the center (0, 5), and the radius is 2,
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which means we can go up 2, down 2, right 2, and left 2
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to get four points on the graph.
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I'm going to go up 2, I'm going to go down 2,
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I'm going to to right 2, and I'm going to go left 2.
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I know that these are four points on the graph.
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Draw my circle, so here's the circle.
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Now we can easily, just from the graph, list the intercepts.
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What's the x-intercept?
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Well, you can see that the graph never crosses the x-axis,
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so there are no x-intercepts.
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What are the y-intercepts?
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Well, we can easily find them from the graph.
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Since the center is on the y-axis, which is (0, 5),
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and we know that the radius is 2,
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so, if we go up 2, this is going to be 7,
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and, if we go down 2, this is going to be 3,
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so the y-intercepts are 3 and 7.
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What's important in this example is that you have to get it in standard form first.
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If there's any number here in front of the x and the y squared,
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you want to divide by that number first.
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Now let's look at another similar example.
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Let's look at this example where we want to find the center and the radius
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and then find the intercepts.
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Notice that this particular equation is in general form,
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and when it's in general form,
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we don't know what the center and the radius are,
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so we're going to have to work backwards.
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Remember we multiplied it out to get from standard form to general form.
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To go from general form to standard form,
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we're going to have to complete the square.
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What I'm going to do is put the x terms together, 4x,
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and then put the y terms together.
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I'm going to bring the constant to the other side.
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Now remember when completing the square,
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what you do is you take half the middle term,
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so take half of 4, which is 2, and you square it.
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I want to add 4 here.
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Again, this is negative 4.
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Half of negative 4 is negative 2,
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but, when you square it, you still get 4.
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Notice what I've added on the lefthand side is 8,
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so, to make the equation still hold true,
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I need to add 8 to both sides.
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I'm going to come over here and add 8.
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Now what I have are two perfect squares.
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This is the perfect square of x plus 2.
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This is the perfect square of y minus 2.
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On the righthand side, we have 9.
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Now that it's in standard form, we can easily find the center,
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which is going to be (-2, 2), and the radius, the square root of 9, is 3.
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Now let's find the intercepts.
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Here's the standard form of the equation of the circle that we got,
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and we're going to find the intercepts using this equation.
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Remember to find the x-intercepts if there are any.
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You're going to set y equal to 0.
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When I set y equal to 0, I'm going to get 0 minus 2.
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On that term, you're going to get x plus 2 squared.
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Since y is 0, you get negative 2 squared,
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which is 4, equals 9. Subtracting 4 from both sides, you get 5.
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Now I'm going to solve this quadratic equation by using the square root method.
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Remember your plus or minus, so I get x plus 2
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equals plus or minus the square root of 5,
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so x equals negative 2 plus or minus the square root of 5.
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Here are your two x-intercepts that you can see are irrational.
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Now let's see if there are any y-intercepts.
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To do that, we're going to set x equal to 0.
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Here, if we set x to 0, we're going to get 4
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plus y minus 2 squared equals 9.
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Again, we're going to subtract 4 from both sides. We get 5.
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Again taking the square root of both sides,
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you get plus or minus the square root of 5.
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Then adding 2 to both sides,
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I get y equals 2 plus or minus the square root of 5.
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Here are two y-intercepts for this particular circle.
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Now let's look at another example.
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Alright, let's look at this example.
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We want to find the general form of the equation of a circle
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given the center (-3, 1),
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and this circle is tangent to the y-axis.
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Remember, what do we need to get the equation of a circle?
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We need to find the center, and we need the radius.
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Well, we have the center, so we need to find the radius.
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Well, let's draw what we know so far.
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We know the center is (-3, 1).
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If the circle is tangent to the y-axis,
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what that means is the circle touches the y-axis at exactly one point,
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which means it's going to touch right here.
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It can't touch anywhere else or it will intersect at two points,
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so that means that the circle is like this.
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It's tangent to the y-axis. It only touches at one point,
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so we can easily find this radius by counting.
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The radius is of course going to be 3 units.
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Now that we know the center and the radius,
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we can get the equation of the circle.
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We're going to start with the standard form,
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which is x minus h squared plus y minus k squared equals r squared.
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We're going to plug in our h and k (-3, 1).
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We're going to get x plus 3 squared plus y minus 1 squared.
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Since the radius is 3, 3 squared gives us 9.
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Now that's the standard form.
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Now to get the general form, we're going to multiply this all out.
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We get x squared plus 6x plus 9
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plus y squared minus 2y plus 1 equals 9.
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Again in general form, you want the x squared term first,
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then the y squared term, then the x term,
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then the y term, and then we're going to collect our constants together.
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We have 10 on the left minus 9 gives us a plus 1 equals 0.
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Here is the general form of the equation of the circle.
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Okay, let's look at another example.
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Alright, let's look at this example.
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We want the general form of the equation of the circle.
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This time, the information that we're given
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is we're given two endpoints of a diameter of a circle.
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Well, let me start off by drawing what we know so far.
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I'm going to plot each of these points.
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If (4, 3) and (0, 1) are two endpoints of a diameter,
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the circle is something like this. Which means this is your diameter.
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And what do you know?
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The center is on every diameter, so am I going to find the center?
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I'm going to find the midpoint of the segment
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of these two points: (4, 3) and (0, 1).
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Remember how to find the midpoint.
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You're going to find the midpoint, which is my center.
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To find the midpoint, you take the x-coordinates,
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you add them up, and you divide by 2.
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We're going to take 4 plus 0
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divided by 2 and add the y-coordinates together.
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Divide by 2. What do we get?
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We're going to get 4 over 2. We get (2, 2) as the center.
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Now we can see from the graph that that seems reasonable
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just by plotting the points. That center seems reasonable.
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We're doing good. We've got the center.
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What else do we need to find the equation of the circle?
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Right, we need the radius.
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We need to find the distance from this center
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out to either point. It doesn't matter which point we use.
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I'm going to use the point (0, 1) with (2, 2).
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The distance between these two points is going to be my radius.
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Using the distance formula, I'm going to do 2 minus 0 squared
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plus 2 minus 1 squared. Let's see what we get. We get 4 plus 1.
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The radius is the square root of 5.
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Okay, now we know the center: (2, 2).
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We know the radius,
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so we can get the standard form of the equation of the circle.
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We're going to do x minus 2 squared plus y minus 2 squared.
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When you square the square root of 5, you get 5.
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Now to get the general form,
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we need to multiply all this out.
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Let's see. We're going to get x squared minus 4x plus 4
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plus y squared minus 4y plus 4
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equals 5. Again, get the x squared term first,
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then the y squared term, then the x term,
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then the y term, and, again collecting the constants,
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I get 8 minus 5, which is 3.
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Here's the general form of the equation of the circle
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with center and with these two endpoints of a diameter.
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Now circles are very important.
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You need to be able to recognize when you see the equation of a circle
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that it's a circle in either form,
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and you need to be able to get the equation given information.
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You need to practice these skills.