What is (1+y)^2+x^2=4^2 ?

The solution to (1+y)^2+x^2=4^2 is Circle with (a,b)=(0,-1),r=4

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(1+y)^2+x^2=4^2

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

(a, b) = (0, - 1), r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 1))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (0, - 1), r = 4

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

x^{2} + y^{2} + 42 x + 84 y + 65 = 0

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

x^{2} - 14 x + y^{2} + 12 y = 36

x^{2} + y^{2} - 8 x - 12 y + 48 = 0

What is (1+y)^2+x^2=4^2 ?
The solution to (1+y)^2+x^2=4^2 is Circle with (a,b)=(0,-1),r=4