Determine the center & radius..please help!?

rewriting x^2=9-(y-3)^2 as

x^2 + (y-3)^2 = 9

x^2 + (y-3)^2 = 3^2

now compare with

(x-h)^2 + (y-k)^2 = r^2

we see that h = 0, k = 3 and r = 3

therefore center is (0,3) and radius is 3

x^2 = 9 - (y - 3)^2

x^2 + (y - 3)^2 = 9

Remember (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center and r is the radius length.

Well, isn't x^2 is also (x - 0)^2?

(h,k) = (0,3) <------------ center

What is the square root of 9? 3?

r = 3 units <------------- radius

Center: (0,3)

Radius: 3

Equation for a circle (x-a)^2+(y-b)^2=r^2 the place the coordinates of the midsection (a,b) radius r. on your eqn. 9x2 + 54x + 9y2 ? 6y + fifty 5 = 0 =>9(x^2+6x)+9(y^2-2/3y)+fifty 5=0 =>9(x^2+6x+9)+9(y^2-2/3y+a million/9)+fifty 5-eighty one-a million=0 =>9(x+3)^2+9(y-2/3)^2=27 =>(x+3)^2+(y-2/3)^2=3 =>(x+3)^2+(y-2/3)^2=(?3)^2 O(-3,+2/3); r=?3 gadgets.

Equation for a circle (x-a)^2+(y-b)^2=r^2 the place the coordinates of the middle (a,b) radius r. on your eqn. 9x2 + 54x + 9y2 ? 6y + fifty 5 = 0 =>9(x^2+6x)+9(y^2-2/3y)+fifty 5=0 =>9(x^2+6x+9)+9(y^2-2/3y+a million/9)+fifty 5-80 one-a million=0 =>9(x+3)^2+9(y-2/3)^2=27 =>(x+3)^2+(y-2/3)^2=3 =>(x+3)^2+(y-2/3)^2=(?3)^2 O(-3,+2/3); r=?3 instruments.