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What is the center and radius of x2 plus 6x plus y2-6y-232 equals 0?
Wiki User
∙ 11y ago
As I understood your question:
"What is the center and radius of: x2+6x+y2-6y-232=0?"
First move 232 over to the right hand side to get: x2+6x+y2-6y = 232.
Complete the square for both x's and y's: (x2 + 6x+ 9) + (y2 - 6y + 9) =232 + 9 + 9
Then rewrite: (x+3)2+(y-3)2=250
Now you can pick off the center --> (-3,3); and
radius --> sqrt(250) = 5sqrt(10) approx. 15.8114
Wiki User
∙ 11y ago
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