What is (x^2-6x)+(y^2+10y)+18=0 ?

The solution to (x^2-6x)+(y^2+10y)+18=0 is Circle with (a,b)=(3,-5),r=4

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(x^2-6x)+(y^2+10y)+18=0

(x^{2} - 6 x) + (y^{2} + 10 y) + 18 = 0

(a, b) = (3, - 5), r = 4

Solution steps

x^{2} - 6 x + y^{2} + 10 y + 18 = 0

Rewrite

x^{2} - 6 x + y^{2} + 10 y + 18 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 5))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (3, - 5), r = 4

(x + 6)^{2} + y^{2} = 25

x^{2} + (y - 2)^{2} = 28

x^{2} + y^{2} - 8 x + 6 y + 21 = 0

1 - x^{2} - y^{2} \geq 0

(x - 6)^{2} + (y - 4)^{2} - 64 = 0

What is (x^2-6x)+(y^2+10y)+18=0 ?
The solution to (x^2-6x)+(y^2+10y)+18=0 is Circle with (a,b)=(3,-5),r=4