What is the answer to solvefor y,x=sqrt(1-y^2) ?

The answer to solvefor y,x=sqrt(1-y^2) is y=sqrt(-x^2+1) {0<= x<= 1},y=-sqrt(-x^2+1) {0<= x<= 1}

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solvefor y,x=sqrt(1-y^2)

solve for

y, x = 1 - y^{2}

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

Solution steps

x = 1 - y^{2}

Square both sides:

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

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

Solve

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

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

Verify Solutions:

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

The solutions are

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

2 x + 7 = x + 2

4 - x^{2} = 0

x = 3

16 - x^{2} = 0

x + 10 = x - 2

What is the answer to solvefor y,x=sqrt(1-y^2) ?
The answer to solvefor y,x=sqrt(1-y^2) is y=sqrt(-x^2+1) {0<= x<= 1},y=-sqrt(-x^2+1) {0<= x<= 1}