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

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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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2%2B2y-12x%2B4y-15=0......here you missing y%5E2 to have a circle
I guess 2y should be 2y%5E2, but it also might be 4y%5E2
let's do it with 2y%5E2
2x%5E2%2B2y%5E2-12x%2B4y-15=0....write it in the standard form
The standard equation of a circle with center C(h,k) and radius r is as follows:
%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
2x%5E2%2B2y%5E2-12x%2B4y-15=0..put all terms with x and x%5E2 together and all terms with y and y%5E2+ together using brackets
%282x%5E2-12x%29%2B%282y%5E2%2B4y%29-15=0......complete the square within each bracket
%282x%5E2-12x%2B18%29-18%2B%282y%5E2%2B4y%2B2%29-2-15=0
2%28x%5E2-6x%2B9%29%2B2%28y%2B2x%2B1%29-35=0
2%28x-3%29%5E2%2B2%28y%2B1%29%5E2-35=0
2%28x-3%29%5E2%2B2%28y%2B1%29%5E2=35 ....both sides divide by 2

2%28x-3%29%5E2%2F2%2B2%28y%2B1%29%5E2%2F2=35%2F2
%28x-3%29%5E2%2B%28y%2B1%29%5E2=17.5

so,the center and radius of the circle are:
C(3,-1)
r=4.18