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

The solution to x^2+4x+y^2-2y-4=0 is Circle with (a,b)=(-2,1),r=3

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

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

(a, b) = (- 2, 1), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 1)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- 2, 1), r = 3

x^{2} + y^{2} + 6 x + 4 y + 12 = 0

(x + 4)^{2} + (y + 12)^{2} - 16 = 0

x^{2} + y^{2} + 6 x + 16 y - 8 = 0

4 x^{2} + 4 y^{2} + 8 x - 4 y - 3 = 0

(x - \frac{1}{2 ( \frac{3}{2} )})^{2} + y^{2} = (\frac{1}{2 ( \frac{3}{2} )})^{2}

What is x^2+4x+y^2-2y-4=0 ?
The solution to x^2+4x+y^2-2y-4=0 is Circle with (a,b)=(-2,1),r=3