What is x^2+y^2+8x-4y+19=0 ?

The solution to x^2+y^2+8x-4y+19=0 is Circle with (a,b)=(-4,2),r=1

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x^2+y^2+8x-4y+19=0

x^{2} + y^{2} + 8 x - 4 y + 19 = 0

(a, b) = (- 4, 2), r = 1

Solution steps

x^{2} + y^{2} + 8 x - 4 y + 19 = 0

Rewrite

x^{2} + y^{2} + 8 x - 4 y + 19 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, 2), r = 1

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

2 x^{2} + 2 y^{2} + 10 x - 2 y = 7

(x - (- 5))^{2} + (y - 7)^{2} = (11)^{2}

x^{2} + y^{2} + 10 x - 8 y + 36 = 0

(x - 1)^{2} + (y + 11)^{2} + (0 - 8)^{2} = 100

What is x^2+y^2+8x-4y+19=0 ?
The solution to x^2+y^2+8x-4y+19=0 is Circle with (a,b)=(-4,2),r=1