SOLUTION: What is the answer in standard form and the center and radius of this circle equation? 3x^2+3y^2-18x+6y=18

Question 139392This question is from textbook Algebra 2
: What is the answer in standard form and the center and radius of this circle equation?
3x^2+3y^2-18x+6y=18 This question is from textbook Algebra 2

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You mean the 'equation' in standard form, I think.

The standard form of an equation of a circle with center at (h,k) and radius r is

Notice that the terms on the left are each a squared binomial, so the trick to converting an equation in the form into standard form is to complete the square on each of the variables.

Notice that all of the coefficients on this one are divisible by 3, so divide through by 3 first because we need the coefficients on the 2nd degree terms to be 1.

is an equivalent equation and describes the same conic section.

Next rearrange the terms so that the x terms are together and the y terms are together:

Divide the coefficient on the 1st degree x term by 2, square the result, and add that result to both sides of the equation:

Do the same thing with the coefficient on the 1st degree y term:
, and collect terms on the right

Now we have a sum of two perfect square trinomials: and , so:

, but so:

Now compare this result to the standard form . We need to make one little adjustment to make our result look exactly like the standard form pattern:

Now we can read the coordinates of the center and the value of the radius directly.

, , so the center is at (3,-1)

and the radius is

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