What is x^2+y^2-12x+4y-140=0 ?

The solution to x^2+y^2-12x+4y-140=0 is Circle with (a,b)=(6,-2),r=6sqrt(5)

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x^2+y^2-12x+4y-140=0

x^{2} + y^{2} - 12 x + 4 y - 140 = 0

(a, b) = (6, - 2), r = 65

Solution steps

x^{2} + y^{2} - 12 x + 4 y - 140 = 0

Rewrite

x^{2} + y^{2} - 12 x + 4 y - 140 = 0

in the form of the standard circle equation

(x - 6)^{2} + (y - (- 2))^{2} = (65)^{2}

Therefore the circle properties are:

(a, b) = (6, - 2), r = 65

x^{2} + y^{2} + 2 x + 2 y = 7

x^{2} - 10 x + y^{2} + 8 = 0

x^{2} + y^{2} - 4 x + 18 y + 35 = 0

- 4 y^{2} - 4 x^{2} - 20 x + 32 y + 81 = 0

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

What is x^2+y^2-12x+4y-140=0 ?
The solution to x^2+y^2-12x+4y-140=0 is Circle with (a,b)=(6,-2),r=6sqrt(5)