SOLUTION: What is the centre and radius of the circle with the equation 2x^2+2y-12x+4y-15=0

Question 730930: What is the centre and radius of the circle with the equation
2x^2+2y-12x+4y-15=0
Found 2 solutions by lwsshak3, MathLover1:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
What is the centre and radius of the circle with the equation
2x^2+2y-12x+4y-15=0
***
Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius
For given equation:
2x^2+2y-12x+4y-15=0
This is not an equation of a circle because the y^2 term is missing.
Looks like a typing error, one of the y-terms should probably be squared.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
......here you missing to have a circle
I guess should be , but it also might be
let's do it with
....write it in the standard form
The standard equation of a circle with center C(,) and radius is as follows:

..put all terms with and together and all terms with and together using brackets
......complete the square within each bracket

....both sides divide by

so,the center and radius of the circle are:
C(,)

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