What is (x+12)^2+(y-4)^2-30=0 ?

The solution to (x+12)^2+(y-4)^2-30=0 is Circle with (a,b)=(-12,4),r=sqrt(30)

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(x+12)^2+(y-4)^2-30=0

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

(a, b) = (- 12, 4), r = 30

Solution steps

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

Rewrite

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

in the form of the standard circle equation

(x - (- 12))^{2} + (y - 4)^{2} = (30)^{2}

Therefore the circle properties are:

(a, b) = (- 12, 4), r = 30

6 x^{2} + 6 y^{2} - 24 x + 36 y = 72

x^{2} + y^{2} - 2 (x + y) > 0

2 x^{2} + 2 y^{2} + 6 x - 8 y + 12 = 0

x^{2} + y^{2} + 8 x - 6 y - 15 = 0

16 x^{2} + 16 y^{2} + 8 x - 32 y = - 1

What is (x+12)^2+(y-4)^2-30=0 ?
The solution to (x+12)^2+(y-4)^2-30=0 is Circle with (a,b)=(-12,4),r=sqrt(30)