What is sqrt(25)^2=(x-0)^2+(y-(-0))^2 ?

The solution to sqrt(25)^2=(x-0)^2+(y-(-0))^2 is Circle with (a,b)=(0,0),r=5

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sqrt(25)^2=(x-0)^2+(y-(-0))^2

25^{2} = (x - 0)^{2} + (y - (- 0))^{2}

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

Solution steps

(25)^{2} = (x - 0)^{2} + (y - (- 0))^{2}

Rewrite

25^{2} = (x - 0)^{2} + (y - (- 0))^{2}

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 14 x - 8 y + 1 = 0

(x - 3)^{2} + (y - 4)^{2} = 5^{2}

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

x^{2} + y^{2} + 14 x - 8 y + 16 = 0

- 21 x - 42.4 = (x - 0)^{2} + (y - 1.6)^{2} - 2. 1^{2}

What is sqrt(25)^2=(x-0)^2+(y-(-0))^2 ?
The solution to sqrt(25)^2=(x-0)^2+(y-(-0))^2 is Circle with (a,b)=(0,0),r=5