What is x^2-6x+y^2-7=0 ?

The solution to x^2-6x+y^2-7=0 is Circle with (a,b)=(3,0),r=4

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x^2-6x+y^2-7=0

x^{2} - 6 x + y^{2} - 7 = 0

(a, b) = (3, 0), r = 4

Solution steps

x^{2} - 6 x + y^{2} - 7 = 0

Rewrite

x^{2} - 6 x + y^{2} - 7 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - 0)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (3, 0), r = 4

x^{2} + y^{2} - 9 x + 4 y - 127 = 0

(x - 5)^{2} + (y - 4)^{2} = 36

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}

x^{2} + y^{2} - 6 x - 18 y + 41 = 0

(x - 1)^{2} + (y - 2)^{2} = 3^{2}

What is x^2-6x+y^2-7=0 ?
The solution to x^2-6x+y^2-7=0 is Circle with (a,b)=(3,0),r=4