What is x^2+y^2-12x+18y-4=0 ?

The solution to x^2+y^2-12x+18y-4=0 is Circle with (a,b)=(6,-9),r=11

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x^2+y^2-12x+18y-4=0

x^{2} + y^{2} - 12 x + 18 y - 4 = 0

(a, b) = (6, - 9), r = 11

Solution steps

x^{2} + y^{2} - 12 x + 18 y - 4 = 0

Rewrite

x^{2} + y^{2} - 12 x + 18 y - 4 = 0

in the form of the standard circle equation

(x - 6)^{2} + (y - (- 9))^{2} = 1 1^{2}

Therefore the circle properties are:

(a, b) = (6, - 9), r = 11

(x - 56)^{2} + (y + 45)^{2} = 81

x^{2} + y^{2} - 8 x + 4 y = 5

x^{2} + y^{2} - x - 6 y + 9 = 0

x^{2} + y^{2} - x - 8 y - 15 = 0

x^{2} + (y - 2.5)^{2} = 6.25

What is x^2+y^2-12x+18y-4=0 ?
The solution to x^2+y^2-12x+18y-4=0 is Circle with (a,b)=(6,-9),r=11