SOLUTION: Find the center, radius and intercepts of the circle with the given equations. 3x^2+3y^2-18x+24y+27 =0

Question 714398: Find the center, radius and intercepts of the circle with the given equations. 3x^2+3y^2-18x+24y+27 =0
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
--> (dividing both sides of the equal sign by 3)
Look at the terms in x,
Since is part of we can "complete the square"
with the already present at the end of the equation.
Now, look at the terms in y,
is part of ,
so adding to both sides of the equal sign in we can rearrange and group to get
-->-->-->
That is the equation of a circle of radius centered at (3,-4),
the point with and
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