SOLUTION: x^2+y^2+8x-6y=0 find the center and radius

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Question 599205: x^2+y^2+8x-6y=0
find the center and radius
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--
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The equation for a circle has the form
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
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The ordered pair (h,k) is the center of the circle and r is the radius. We need to transform the given equation to this form to find the center and radius.
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x%5E2%2By%5E2%2B8x-6y=0
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We will use the "completing the square" method. The polynomial x^2+8x+16 is a perfect square [(8/2)^2=16], so we need to add 16 to both sides of the equation and re-arrange the terms.
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%28x%5E2%2B8x%2B16%29%2By%5E2-6y=0%2B16
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The polynomial y^2-6y+9 is a perfect square, so we need to add 9 to both sides of the equation.
%28x%5E2%2B8x%2B16%29%2B%28y%5E2-6y%2B9%29=16%2B9
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Now write the polynomial expressions in factored form.
%28x%2B4%29%5E2%2B%28y-3%29%5E2=25
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Then write the constant term as the square of a number.
%28x%2B4%29%5E2%2B%28y-3%29%5E2=5%5E2
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The equation is now in standard form, so the center=(h,k) is (-4,3), and the radius is 5.
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That's it. Feel free to email via my gmail account if you have questions about the solution.
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Ms.Figgy
math.in.the.vortex@gmail.com