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What is the standard form of the equation of the circle x^2 - 2x + y^2 - 8y + 8 = 0?

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Answer:

Explanation:

Begin by grouping the x terms and the y terms together and separating the constants out.


(x^2-2x)+(y^2-8y)=-8

Now we'll complete the square on those x and y terms. Take half the linear term of each, square it, and add it to both sides. Our linear x term is 2, half of 2 is 1 and 1 squared is 1, so we add that in. Likewise, half the linear y term (which is 8) is 4, and 4 squared is 16, so we add that in, too. Like this:


(x^2-2x+1)+(y^2-8x+16)=-8+1+16

Doing this gives us the perfect square binomials for each of the x and y terms, and then gives us the radius on the right:


(x-1)^2+(y-4)^2=9

This is a circle with a center of (1, 4) and a radius of 3.

User Jayr Motta
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