Question 110005
Given the equation:
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{{{x^2+y^2-4x-6y-5=0}}}
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We need to get this into the standard form for the equation of a circle. Begin by getting rid
of the -5 on the left side. You can do this by adding +5 to both sides so that you get:
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{{{x^2 + y^2 -4x - 6y = 5}}}
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Next collect the x terms and the y terms on the left side. When you do that the equation becomes:
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{{{(x^2 - 4x)+(y^2 - 6y) = 5}}}
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Next complete the square for each of the two quantities inside the parentheses. For the first
quantity you will complete the square of {{{x^2 - 4x}}}. Because the coefficient (multiplier) of
the {{{x^2}}} term is one, you can find what has to be added to complete the square by using
the following method. Divide the multiplier of the x term by 2 and square the result. This
means that you divide -4 by 2 and get -2. Then square the -2 to get +4. So you need to add 
+4 inside the parentheses. But if you add +4 to the left side of the equation, you must also
add +4 to the right side. This means the equation becomes:
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{{{(x^2 - 4x + 4)+(y^2 - 6y) = 5+ 4}}}
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The polynomial inside the first set of parentheses is now a perfect square, and you can write
it as such. On the right side the 5 and the +4 can be added to get 9.  This makes the 
equation become:
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{{{(x -2)^2 + (y^2 - 6y) = 9}}}
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Next you do the same thing for the y-terms in the second set of parentheses. Since the
coefficient of the y^2 term is 1, you can find the number you need to add in this second
set of parentheses to form a perfect square. Take half the coefficient of the y term, that is,
take half of -6 to get -3 and square this to get +9. Therefore, by adding 9 inside these
parentheses you get a perfect square. And if you add 9 on the left side you must also add 9 on 
the right side of the equation. Doing both of these actions results in:
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{{{(x-2)^2 + (y^2 - 6y + 9) = 9 + 9}}}
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Next write the polynomial in the second set of parentheses as a perfect square and also add the 
two 9s on the right side to get 18. When you do these two things, the equation becomes:
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{{{(x-2)^2 + (y - 3)^2 = 18}}}
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This is the standard form of an equation of a circle. In this case, the center of the circle
is located at the point (2, 3) and the radius of the circle is the square root of the right
side ... or {{{sqrt(18)}}}.
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The radius can be simplified by applying some laws of square roots. In particular:
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{{{sqrt(18) = sqrt(9*2) = sqrt(9)*sqrt(2) = 3*sqrt(2)}}}
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So that is the answer ... the radius of the circle is {{{3*sqrt(2)}}}
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Hope this is understandable to you. It will help if you relate this to an explanation in
a textbook so you get more of a feel for what is going on ... Please verify the math to 
make sure that I didn't make a sign error or some other simple mistake in doing this work.
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