SOLUTION: Graph the circle {{{x^2+y^2-4x+10y+20=0}}}

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Question 244978: Graph the circle x%5E2%2By%5E2-4x%2B10y%2B20=0
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
to graph this circle, you need to solve for y.

solving for y gets you:

y = +/- sqrt (-x^2+4x+5) - 5

graph looks like this:



the center of this circle looks like it might be (x,y) = (2,-5).

in order to know for sure, we have to transform the equation into the proper form.

the proper form is (x-h)^2 + (y-k)^2 = r^2
where (h,k) is the center of the circle.

the original equation is:

x^2+y^2-4x+10y+20=0

move the terms around until you have all the x's together and all the y's together.

move the constant term to the right side of the equation.

you get:

(x^2 - 4x) + (y^2 + 10y) = -20

complete the squares for both of these to get:

(x-2)^2 + (y+5)^2 = -20 + 4 + 25

this becomes:

(x-2)^2 + (y+5)^2 = 9

The center of the circle is at (x,y) = (2,-5) and the radius of the circle is 3.

the graph confirms that.