What is (x+2)^2+(y-8)^2=4^2 ?

The solution to (x+2)^2+(y-8)^2=4^2 is Circle with (a,b)=(-2,8),r=4

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(x+2)^2+(y-8)^2=4^2

(x + 2)^{2} + (y - 8)^{2} = 4^{2}

(a, b) = (- 2, 8), r = 4

Solution steps

(x + 2)^{2} + (y - 8)^{2} = 4^{2}

Rewrite

(x + 2)^{2} + (y - 8)^{2} = 4^{2}

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 8)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 2, 8), r = 4

(x - 40)^{2} + (y - 0)^{2} = 0.1

3 x^{2} - 2 x + 3 + 3 y^{2} + 8 y + 48 = \frac{172}{3}

(x - 1)^{2} + (y + 5)^{2} < 4

(X + 4)^{2} + (Y - 1)^{2} = 148

(x + 2)^{2} + (y + 3)^{2} = 11

What is (x+2)^2+(y-8)^2=4^2 ?
The solution to (x+2)^2+(y-8)^2=4^2 is Circle with (a,b)=(-2,8),r=4