What is (x-(-5))^2+(y-0)^2=4^2 ?

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

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

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

(a, b) = (- 5, 0), r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 5, 0), r = 4

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

(x + \frac{19}{2})^{2} + (y - 230)^{2} = 9

(x + 1)^{2} + (y + 4)^{2} = 16

(x + 5)^{2} + y^{2} = 45

(x - 5)^{2} + (y - 5)^{2} = 25

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