What is x^2+10x+y^2-4-8=12 ?

The solution to x^2+10x+y^2-4-8=12 is Circle with (a,b)=(-5,0),r=7

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x^2+10x+y^2-4-8=12

x^{2} + 10 x + y^{2} - 4 - 8 = 12

(a, b) = (- 5, 0), r = 7

Solution steps

x^{2} + 10 x + y^{2} - 4 - 8 = 12

Rewrite

x^{2} + 10 x + y^{2} - 4 - 8 = 12

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 0)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (- 5, 0), r = 7

4 (x + 1)^{2} + 4 (y)^{2} = 121

(x - 5)^{2} + (y + 3)^{2} = 73

(x + 4)^{2} + (y - 3)^{2} = 16

(x - 5)^{2} + (y + 3)^{2} = 85

(x - 1)^{2} + (y - 4)^{2} = 12

What is x^2+10x+y^2-4-8=12 ?
The solution to x^2+10x+y^2-4-8=12 is Circle with (a,b)=(-5,0),r=7