What is (x-4)^2+(y-20)^2=sqrt(592)^2 ?

The solution to (x-4)^2+(y-20)^2=sqrt(592)^2 is Circle with (a,b)=(4,20),r=sqrt((\sqrt{592))^2}

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(x-4)^2+(y-20)^2=sqrt(592)^2

(x - 4)^{2} + (y - 20)^{2} = 592^{2}

(a, b) = (4, 20), r = (592)^{2}

Solution steps

(x - 4)^{2} + (y - 20)^{2} = (592)^{2}

Rewrite

(x - 4)^{2} + (y - 20)^{2} = 592^{2}

in the form of the standard circle equation

(x - 4)^{2} + (y - 20)^{2} = ((592)^{2})^{2}

Therefore the circle properties are:

(a, b) = (4, 20), r = (592)^{2}

x^{2} + x + y^{2} + \frac{8}{3} y - \frac{1223}{36} = 0

(x - 4)^{2} + (y - 2)^{2} = 36

x^{2} + y^{2} + 2 x - 10 y + 1 = 0

(x - 2)^{2} + (y + 8)^{2} = 144

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

What is (x-4)^2+(y-20)^2=sqrt(592)^2 ?
The solution to (x-4)^2+(y-20)^2=sqrt(592)^2 is Circle with (a,b)=(4,20),r=sqrt((\sqrt{592))^2}