What is the answer to solvefor y,sqrt(x^2+y^2)=x^2+y^2 ?

The answer to solvefor y,sqrt(x^2+y^2)=x^2+y^2 is

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solvefor y,sqrt(x^2+y^2)=x^2+y^2

solve for

y, x^{2} + y^{2} = x^{2} + y^{2}

y = - x^{2} + 1 {- 1 \leq x \leq 1}, y = - - x^{2} + 1 {- 1 \leq x \leq 1}, y = - x^{2} {x = 0}, y = - - x^{2} {x = 0}

Solution steps

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

Square both sides:

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

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

Solve

x^{2} + y^{2} = x^{4} + 2 y^{2} x^{2} + y^{4} : y = - x^{2} + 1, y = - - x^{2} + 1, y = - x^{2}, y = - - x^{2}

y = - x^{2} + 1, y = - - x^{2} + 1, y = - x^{2}, y = - - x^{2}

Verify Solutions:

y = - x^{2} + 1 {- 1 \leq x \leq 1}, y = - - x^{2} + 1 {- 1 \leq x \leq 1}, y = - x^{2} {x = 0}, y = - - x^{2} {x = 0}

The solutions are

y = - x^{2} + 1 {- 1 \leq x \leq 1}, y = - - x^{2} + 1 {- 1 \leq x \leq 1}, y = - x^{2} {x = 0}, y = - - x^{2} {x = 0}

- 1 = 6 x + 21 - x

2 x + 1 - 3 4 x - 3 - 5 = 0

x - 2 x - 1 = 1 - x

x^{2} - 5 + x - 2 = 3

y, x y = x - 2 y

What is the answer to solvefor y,sqrt(x^2+y^2)=x^2+y^2 ?
The answer to solvefor y,sqrt(x^2+y^2)=x^2+y^2 is