What is x^{(2)}+y^{(2)}+4x-4y-1=0 ?

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

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x^{(2)}+y^{(2)}+4x-4y-1=0

x^{(2)} + y^{(2)} + 4 x - 4 y - 1 = 0

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

Solution steps

x^{2} + y^{2} + 4 x - 4 y - 1 = 0

Rewrite

x^{2} + y^{2} + 4 x - 4 y - 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x, x^{2} = 8 - r (x)^{2}

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

x^{2} + y^{2} - 10 x + 4 y - 20 = 0

x^{2} + y^{2} - 12 x + 14 y + 4 = 0

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

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