What is X^2-2X+Y^2+8Y+4=0 ?

The solution to X^2-2X+Y^2+8Y+4=0 is Circle with (a,b)=(1,-4),r=sqrt(13)

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X^2-2X+Y^2+8Y+4=0

X^{2} - 2 X + Y^{2} + 8 Y + 4 = 0

(a, b) = (1, - 4), r = 13

Solution steps

X^{2} - 2 X + Y^{2} + 8 Y + 4 = 0

Rewrite

X^{2} - 2 X + Y^{2} + 8 Y + 4 = 0

in the form of the standard circle equation

(X - 1)^{2} + (Y - (- 4))^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (1, - 4), r = 13

(x - 3)^{2} + (y + 5)^{2} = 106

x^{2} + 8 y + y^{2} + 2 y + 8 = 0

x^{2} + y^{2} - 16 x - 12 y + 75 = 0

x^{2} + (y - 2.25)^{2} = \frac{196}{169}

x^{2} - 6 x + y^{2} - 6 y + 11 = 0

What is X^2-2X+Y^2+8Y+4=0 ?
The solution to X^2-2X+Y^2+8Y+4=0 is Circle with (a,b)=(1,-4),r=sqrt(13)