What is (5-x)^2+z^2=4 ?

The solution to (5-x)^2+z^2=4 is Circle with (a,b)=(5,0),r=2

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(5-x)^2+z^2=4

(5 - x)^{2} + z^{2} = 4

(a, b) = (5, 0), r = 2

Solution steps

(5 - x)^{2} + z^{2} = 4

Rewrite

(5 - x)^{2} + z^{2} = 4

in the form of the standard circle equation

(x - 5)^{2} + (z - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (5, 0), r = 2

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

(x + \frac{2}{3})^{2} + (y + 2)^{2} = 16

x^{2} + y^{2} + 6 x - 10 y + 25 = 0

x^{2} + y^{2} + 4 x + 2 y - 11 = 0

(x - 2)^{2} + (y + 1)^{2} = 64

What is (5-x)^2+z^2=4 ?
The solution to (5-x)^2+z^2=4 is Circle with (a,b)=(5,0),r=2