What is x^2-x=4-z^2-2z ?

The solution to x^2-x=4-z^2-2z is Circle with (a,b)=(1/2 ,-1),r=(sqrt(21))/2

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x^2-x=4-z^2-2z

x^{2} - x = 4 - z^{2} - 2 z

(a, b) = (\frac{1}{2}, - 1), r = \frac{21}{2}

Solution steps

x^{2} - x = 4 - z^{2} - 2 z

Rewrite

x^{2} - x = 4 - z^{2} - 2 z

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (z - (- 1))^{2} = (\frac{21}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - 1), r = \frac{21}{2}

x^{2} + y^{2} + 2 x - 4 y - 5 = 0

x^{2} + y^{2} - x + 3 y - 5 = 0

x^{2} + y^{2} \geq 16

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

x^{2} + y^{2} - 1 > 0

What is x^2-x=4-z^2-2z ?
The solution to x^2-x=4-z^2-2z is Circle with (a,b)=(1/2 ,-1),r=(sqrt(21))/2