What is x^2-4x-3+y^2+10y-4=0 ?

The solution to x^2-4x-3+y^2+10y-4=0 is Circle with (a,b)=(2,-5),r=6

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x^2-4x-3+y^2+10y-4=0

x^{2} - 4 x - 3 + y^{2} + 10 y - 4 = 0

(a, b) = (2, - 5), r = 6

Solution steps

x^{2} - 4 x - 3 + y^{2} + 10 y - 4 = 0

Rewrite

x^{2} - 4 x - 3 + y^{2} + 10 y - 4 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, - 5), r = 6

x^{2} + y^{2} = 5 3^{2}

x^{2} + y^{2} - 14 x - 18 y + 114 = 0

(x + 1)^{2} + (y - 2)^{2} = 100

x^{2} + y^{2} + 6 x - 2 y - 10 = 0

(x - (- 7))^{2} + (y - (- 6))^{2} = 19^{2}

What is x^2-4x-3+y^2+10y-4=0 ?
The solution to x^2-4x-3+y^2+10y-4=0 is Circle with (a,b)=(2,-5),r=6