WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today we are going to talk about circles, beginning with the definition of a circle.
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A **circle** is defined as the set of points in the plane equidistant from a given point, called the **center**.
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For example, if you had a center of a circle here, and you measured any point's distance from the center, these would all be equal.
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And the radius is the segment with endpoints at the center and at a point on the circle.
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The equation for the circle is given as follows: if the center is at (h,k) and the radius is r,
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then the equation is (x - h)² + (y - k)² = r².
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And this is the standard form; and just as with parabolas, the standard form gives you a lot of useful information
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and allows you to graph what you are trying to graph.
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For example, if I were given (x - 4)² + (y - 5)² = 9, then I would have a lot of information.
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I would know that my center is at (h,k), so it is at (4,5).
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And the radius...r² = 9; therefore, r = √9, which is 3.
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So, based on this information, I could work on graphing out my circle.
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Use symmetry to graph a circle, as well as what you discover from looking at the equation in standard form.
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Looking at a different equation, (x - 1)² + (y - 3)² = 4: this equation describes the circle
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with the center at (h,k), which is (1,3); r² = 4; therefore, r = 2.
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So, I have a circle with a radius of 2 and the center at (1,3).
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So, if this is (1,3) up here, and I know that the radius is 2, I would have a point here; I would have a point up here.
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Symmetry: I know that, if I divide a circle up, I could divide it into four symmetrical quarters, for example.
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So, if I have this graphed, I could use symmetry to find the other three sections of this circle.
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All right, if the center is not at the origin, then we need to use completing the square to get the equation in standard form.
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Remember that standard form of a circle is (x - h)² + (y - k)² = r².
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With parabolas, we put those in standard form by completing the square.
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But at that time, we were just having to complete the square of either the x variable terms or the y variable terms.
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Now, we are going to be working with both; and as always, we need to remember to add the same thing to both sides to keep the equation balanced.
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If I was looking at something such as x² + y² - 4x - 8y - 5 = 0,
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what I am going to do is keep all the x variable and y variable terms together, and then just move the constant over to the right.
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The other thing I am going to do is group the x variable terms together: x² - 4x is grouped together, just like up here.
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And then, I am going to have y² - 8y grouped together, and add 5 to both sides.
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Now, I have to complete the square for both of these.
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This is going to give me x² - 4x, and then here I need to have b²/4.
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Since b is 4, that is going to give me 4²/4 = 16/4 = 4; so, I am going to put a 4 in here.
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For the y expression, I am going to have...let's do this up here...b²/4 = 8²/4, which is going to come out to 16.
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So, I am going to add 16 here; and I need to make sure I do the same thing on the right.
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So, I need to add 4, and I need to add 16; if I don't, this won't end up being balanced.
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Now, I want this in this form; so let's change it to (x - 2)² +...here it is going to be (y - 4)² =...
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4 and 16 is 20, plus 5; so that is 25.
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This gives me the equation in standard form; and the center is at (h,k), (2,4).
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And the radius...well, r² is 25; therefore, the radius equals 5.
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And as always, you need to be careful: let's say I ended up with something in this form, (x + 3)² + (y - 2) = 9.
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The temptation for the center might be just to put (3,2); but standard form says that this should be a negative.
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So, I may even want to rewrite this as (x - (-3))² + (y - 2) = 9, just to make it clear that the center is actually at (-3,2).
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And then, the radius is going to be the square root of 9, which is 3.
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So, be careful that you look at the signs; and if the signs aren't exactly the same as standard form,
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you need to compensate for that, or even just write it out--because -(-3) would give me +3, so these two are interchangeable.
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All right, in this example, we are asked to find the equation of the circle which has a diameter with the endpoints (-3,-7) and (9,-1).
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So, let's see what we are working with.
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Just sketch this out at (-3,-7), right about there; over here is (9,-1); there we have the diameter of the circle.
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We have a circle like this, and we want to find the equation.
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Recall that the formula for the equation of a circle is (x - h)² + (y - k)² = r².
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Therefore, I need to find h; I need to find k; and I need to find the radius.
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Recall that (h,k) gives you the center of the circle.
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Since this is the diameter of the circle, the center of this line segment is going to be the center of the circle.
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So, (-3,-7)...over here I have (9,-1); and here I have the center--the center is going to be equal to the midpoint of this segment.
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Recall the midpoint formula equals (x₁ + x₂)/2, and then (y₁ + y₂)/2.
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So, the center--the coordinates for that are going to be equal to (-3 + 9)/2, and then (-7 + -1)/2,
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which is going to be equal to 6/2...-7 and -1 is -8/2; which is equal to (3,-4).
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This means that h and k are 3 and -4; so I have h and k; I need to find the radius.
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Well, half of the diameter...this is all the diameter; this is my midpoint; and I know that this is at (3,-4).
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So, I just need to find this length--this is the radius.
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And I now have endpoints; so I can use either one of these--I have a set of endpoints here and here, and here and here.
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I am going to go ahead and use these two, and put them into the distance formula.
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(-3,-7) and (3,-4)--I can use these in the distance formula: the distance here equals the radius,
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which is the square root of...I am going to make this (x₁,y₁), and then this (x₂,y₂).
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So, this is going to give me...x₂ is 3, minus -3, squared, plus...y₂ is -4, minus -7, squared.
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So, the radius equals the square root of 3 + 3; a negative and a negative is a positive; all of this squared,
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plus -4...and a negative and a negative is a positive, so -4 + 7, squared.
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So, the radius equals the square root of...3 + 3 gives me 6, squared, plus...7 - 4 gives me 3, squared.
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So, the radius equals √(36 + 9); 36 plus 9 is 45, so the radius equals √45.
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But what I really want for this is r², so r² is going to equal (√45)², which is going to equal 45.
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Putting this all together, I can write my equation, because I now have h; I have k; and I have r.
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So, writing the equation up here gives me (x - 3)² + (y...I can either write this as - -4,
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or I can rewrite this as (y + 4)² = r², which is 45.
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So, this, or a little more neatly, like this: (y + 4)² = 45--this is the equation for the circle.
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And I found that information based on simply knowing the diameter.
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Knowing the diameter, I could use the midpoint formula to find the center, which gave me h and k.
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And then, I could use the distance formula to find the distance from the center to the end of the diameter, which gave me the radius.
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And then, I squared the radius and applied it to that formula.
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Example 2: Find the center and radius of the circle with this equation.
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In order to achieve that, we need to put this equation in standard form.
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And recall that standard form of a circle is (x - h)² + (y - k)² = r².
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So, we need to complete the square: group the x variable terms together on the left; also group the y variable terms on the left.
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Add 8 to both sides to move the constant over.
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I need x² - 8x + something to complete the square, and y² + 10y + something to complete the square, equals 8.
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So, for the x variable terms, I want b²/4, and this is going to be 8²/4, or 64/4, equals 16.
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So, I am going to add 16 here.
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For the y variable terms, b²/4 is going to equal 10²/4, which is 100/4, which is 25.
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I need to be careful that I add the same thing to both sides to keep this equation balanced, so I am also going to add 16 and 10 to the right side.
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Correction: it is 16 and 25--there we go.
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(x - 4)² equals this perfect square trinomial, and I am trying to get it in this form; that is what I want it to look like.
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Plus...(y + 5)² comes out to this perfect square trinomial.
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On the right, if I add 8 and 16 and 25, I am going to end up with 49.
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8 and 16 is going to give me 24, plus 25 is going to give me 49.
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Now, I have this in standard form: because the center of a circle is (h,k), I know I have h here,
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and I know I have k here, this is going to give me...h is 4; k is -5.
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Be careful with the sign here, because notice: this is (y + 5), but standard form is - 5, so this is equal to (y - -5)²--the same thing.
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It is just simpler to write it like this; but make sure you are careful with that.
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The radius here: well, I have r²; the radius, r², I know, is equal to 49.
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Therefore, r = √49, so the radius equals 7.
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Therefore, the center of this circle is at (4,-5), and the radius is equal to 7.
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Example 3: Find the radius of the circle with the center at (-3,-4) and tangent to the y-axis.
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This one takes more drawing and just thinking, versus calculating.
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The center is at (-3,-4), right about here.
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The other thing I know about this circle is that it is tangent to the y-axis.
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That means that, if I drew the circle, it is going to extend around, and it is going to touch this y-axis.
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Well, the radius is going to have one endpoint on the circle, and the other endpoint at the center.
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Therefore, the radius has to extend from (-3,-4) over here.
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And at this point, we are able to then find the length, because, since x is -3, and it has to go all the way to x = 0,
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then this distance must be 3; therefore, the radius equals 3.
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And again, that is because I know the center is here at -3, and I know
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that the other endpoint of the radius is going to be out here, forming the circle,
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and that, because it is tangent to the y-axis, x is going to be equal to 0 right here.
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So, I know that x is equal to 0 here; and I know that x is equal to -3 over here; so it is just 1, 2, 3 over--this distance here is going to be 3.
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And that is going to be the same as the radius, so the radius is equal to 3.
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Example 4: Find the equation of the circle that is tangent to the x-axis, to x = 7, and to x = -5.
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We are given a bunch of information about this circle and told to put it in standard form.
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The first thing we are told is that it is tangent to the x-axis; so this circle is touching the x-axis; let's just draw a line here to emphasize that.
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It is also tangent to x = 7; x = 7 is going to be right here--it is tangent to this.
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It is also tangent to x = -5, out here.
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I am going to end up with a circle that is touching, that is tangent to, these three things.
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Let's think about what that tells me.
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I need to find h and k (I need to find the center).
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I also need to find the radius, so I can find r².
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If this extends from -5 to 7, that gives me the diameter.
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So, the diameter goes from 7 all the way to -5; so if I just add 7 and 5 (the distance from here to here,
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plus the distance from here to here), I am going to get that the diameter equals 12.
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The radius is 1/2 the diameter, so the radius equals 6.
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I found that the radius equals 6.
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The radius is going to extend from these endpoints to the center.
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And I know that it is 6, so I know that the radius is going to go from 7 over here, 6 away from that.
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7 - 6 is 1; it is going to go up to x = 1.
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Again, that is because the length of the radius is 6, so the distance between the center and this endpoint has to be 6.
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7 - 6 is 1; the radius is going to extend from there to there.
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Therefore, the x-value of the center has to be 1.
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Now, what is the y-value of the radius? The other thing I know is that this circle is tangent to this x-axis.
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So, I know that it is going to have an endpoint on the x-axis.
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And then, if it is going to extend from here to here, it is going to have to go up to 6; therefore, the center is at (1,6).
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All right, so the radius equals 6, and the center is at (1,6); and that gives me an equation:
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(x - 1)² + (y - 6)² = the radius squared.
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If r = 6, then r² = 36.
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Again, that is based on knowing that this is tangent to x = -5, x = 7, and the x-axis.
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So, I had the diameter, 12; I divided that by 2 to get the radius; I know that I have an endpoint here and an endpoint at the center.
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So, that gives me the x-value of the center, which is 1.
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I know I have an endpoint here, and also an endpoint at the center; so it has to be up at 6.
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That gives me (1,6) for my value.
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OK, and this is just drawn schematically, because the center would actually be higher up here.
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This is just to give you...the center is actually going to be up here, now that I have my value: it is going to be at (1,6).
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OK, that concludes this lesson of Educator.com on circles; thanks for visiting!