Question

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

Answer 1

Answer:

Step-by-step 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.