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
That's the equation of a circle with its center at the origin and a radius of 8.
Centre = (0,0), radius = sqrt(5)
Center is (0, 0) . . . the origin.Radius is 7.
Centre = (0,0), the origin; radius = 11
The centre is (3,-1) and the radius is sqrt(10).
Draw a circle with its center at the origin and a radius of 3.
It is the equation of a circle with radius of 6 and its center at the origin.
The center of the circle is at (0, 0) and its radius is the square root of 1 which is 1
x2+y2+4x+2y+3=0(x+2)2 + (Y+2.5)2 = 3This is the equation of circle with center at (-2,-2.5) with radius 3.5
Type your answer here. Find the radius for a circle with the equation x2 plus y2 equals 9? ..
The radius will depend on the plus or minus value of 10 or whether or not it needs a plus sign but the center of the circle is at (-2, 3)
You are describing a circle, with its center at the origin and a radius of 4 (the square root of 16)