What is x^2-6x+y^2-32y=0 ?

The solution to x^2-6x+y^2-32y=0 is Circle with (a,b)=(3,16),r=sqrt(265)

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x^2-6x+y^2-32y=0

x^{2} - 6 x + y^{2} - 32 y = 0

(a, b) = (3, 16), r = 265

Solution steps

x^{2} - 6 x + y^{2} - 32 y = 0

Rewrite

x^{2} - 6 x + y^{2} - 32 y = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - 16)^{2} = (265)^{2}

Therefore the circle properties are:

(a, b) = (3, 16), r = 265

(y - 2)^{2} = 8 (x - 1)

\frac{y ^{2}}{9} - \frac{x ^{2}}{4} = 1

y = 3 x^{2}

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1

\frac{( x - 1 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{16} = 1

What is x^2-6x+y^2-32y=0 ?
The solution to x^2-6x+y^2-32y=0 is Circle with (a,b)=(3,16),r=sqrt(265)