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

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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) About Me  (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 %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2

Notice that the terms on the left are each a squared binomial, so the trick to converting an equation in the form Ax%5E2%2BBy%5E2%2BCx%2BDy%2BC=0 into standard form is to complete the square on each of the variables.

3x%5E2%2B3y%5E2-18x%2B6y=18

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.

x%5E2%2By%5E2-6x%2B2y=6 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:

x%5E2-6x%2By%5E2%2B2y=6

Divide the coefficient on the 1st degree x term by 2, square the result, and add that result to both sides of the equation: %28-6%2F2%29%5E2=%28-3%29%5E2=9

x%5E2-6x%2B9%2By%5E2%2B2y=6%2B9

Do the same thing with the coefficient on the 1st degree y term:
%282%2F2%29%5E2=1%5E2=1, and collect terms on the right

x%5E2-6x%2B9%2By%5E2%2B2y%2B1=6%2B9%2B1
x%5E2-6x%2B9%2By%5E2%2B2y%2B1=16

Now we have a sum of two perfect square trinomials: x%5E2-6x%2B9=%28x-3%29%5E2 and y%5E2%2B2y%2B1=%28y%2B1%29%5E2, so:

%28x-3%29%5E2%2B%28y%2B1%29%5E2=16, but 16=4%5E2 so:

%28x-3%29%5E2%2B%28y%2B1%29%5E2=4%5E2

Now compare this result to the standard form %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2. We need to make one little adjustment to make our result look exactly like the standard form pattern:

%28x-3%29%5E2%2B%28y-%28-1%29%29%5E2=4%5E2

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

h=3, k=-1, so the center is at (3,-1)

and the radius is r=4