Question 70950: can you help me understand this problem
find the center and the radius of the circle
x^2+6x+y^2-12y+20=0
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let's start by writing down the standard equation for a circle:
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The center of the circle is (h,k) and r is the radius of the circle. Now what we need to
do is to figure out a way that we can work the equation you were given and get it into the
above form ... minus and plus signs, exponents, parentheses, and everything else in the
same place as the standard circle equation above. Let's begin by writing down the given equation:
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Now let's group the x and y terms:
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Notice that the first group of terms, the (x^2+6x) needs to be a perfect square so we
can convert it to the form (x-h)^2. What we are missing is a constant. If we had it
in the form x^2 + 6x + 9 we would have a perfect square of (x+3)^2. So let's add a 9 to the
terms and we get the equation to be:
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But we can't just add +9 to one side of the equation without adding +9 to the other side
too. So let's put a +9 on the right side also and the equation is now:
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But how did I know to add +9? Simple. I looked at the coefficient of the x term (it is +6),
divided it by two (and got +3), and squared that number (3*3) to get +9. Notice that the
term (x^2 + 6x + 9) is a perfect square. It is equal to (x + 3)^2. [The three comes from
half of the coefficient of the 6x term.] So, in the equation let's now replace the terms
x^2 + 6x + 9 by (x + 3)^2 to get:
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I'm going to do one other small, but necessary change. I'm going to replace the +3 by -(-3)
for reasons we'll see later. So at this point the equation is:
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Now let's do the same thing for the y^2 - 12y. Let's convert it to a perfect square. We
do it in the same way that we did for the x terms. We look at the coefficient of the y
term. It is -12. We divide it by 2 to get -6, an we square that to get +36. We are going
to add +36 in with the y terms, but when we do we also have to add a +36 to the other side
of the equation. Here's the resulting equation:
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Now we can write the y term grouping as a perfect square (y - 6)^2. How do we know that
it's -6? Because -6 is half of the -12 coefficient of y. So let's put this into the equation
for the y term grouping. When we do the equation becomes:
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We've got one more thing to do. Notice on the left side there is a +20 term that doesn't
belong there. Let's get rid of it by subtracting 20 from both sides of the equation. The
result of that subtraction is that we are left with:
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I've written the standard form of the equation of a circle underneath the equation that
we developed. Notice that we can compare them favorably, term by term.
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By comparison we can see that h = -3, k = 6, and r^2 = 25. h and k identify the center of
the circle. The center is the point (h,k) which from our info is (-3,6) and the radius of
the circle is found by taking the square root of both sides of the equation r^2 = 25.
The result of that is r = 5.
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A lot of work, but pretty straightforward. You can follow this same process for problems of
this type.
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I hope that this helps you to see how to progressively work your way through problems that
involve circle equations.
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