What are the solutions to the equation solvefor y,x^2+(y-2)^2=4 ?

The solutions to the equation solvefor y,x^2+(y-2)^2=4 are y=sqrt(4-x^2)+2,y=-sqrt(4-x^2)+2

Find the zeros of solvefor y,x^2+(y-2)^2=4

The zeros of solvefor y,x^2+(y-2)^2=4 are y=sqrt(4-x^2)+2,y=-sqrt(4-x^2)+2

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solve for

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

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

Solution steps

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

Move

x^{2}

to the right side

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

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

5 x^{2} - 50 x + 60 = - 5

- 15 x^{2} + 17 x - 6 = 0

3 x^{2} - 21 x - 54 = 0

(x + 3)^{2} = 14

- 9 x^{2} - 4 x - 2 = 0

What are the solutions to the equation solvefor y,x^2+(y-2)^2=4 ?
The solutions to the equation solvefor y,x^2+(y-2)^2=4 are y=sqrt(4-x^2)+2,y=-sqrt(4-x^2)+2
Find the zeros of solvefor y,x^2+(y-2)^2=4
The zeros of solvefor y,x^2+(y-2)^2=4 are y=sqrt(4-x^2)+2,y=-sqrt(4-x^2)+2