What are the solutions to the equation solvefor y,(x-3)^2+(y+2)^2=9 ?

The solutions to the equation solvefor y,(x-3)^2+(y+2)^2=9 are y=sqrt(-x^2+6x)-2,y=-sqrt(-x^2+6x)-2

Find the zeros of solvefor y,(x-3)^2+(y+2)^2=9

The zeros of solvefor y,(x-3)^2+(y+2)^2=9 are y=sqrt(-x^2+6x)-2,y=-sqrt(-x^2+6x)-2

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solve for

y, (x - 3)^{2} + (y + 2)^{2} = 9

y = - x^{2} + 6 x - 2, y = - - x^{2} + 6 x - 2

Solution steps

(x - 3)^{2} + (y + 2)^{2} = 9

Move

(x - 3)^{2}

to the right side

(y + 2)^{2} = 9 - (x - 3)^{2}

Simplify

9 - (x - 3)^{2} : - x^{2} + 6 x

For

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

the solutions are

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

Solve

y + 2 = - x^{2} + 6 x : y = - x^{2} + 6 x - 2

Solve

y + 2 = - - x^{2} + 6 x : y = - - x^{2} + 6 x - 2

The solutions to the quadratic equation are:

y = - x^{2} + 6 x - 2, y = - - x^{2} + 6 x - 2

3 x^{2} + 9 x = 12

2 t^{2} + 9 t - 3 = 0

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

x^{2} - 12 x + 3 = 0

4 x^{2} - 28 x + 53 = 0

What are the solutions to the equation solvefor y,(x-3)^2+(y+2)^2=9 ?
The solutions to the equation solvefor y,(x-3)^2+(y+2)^2=9 are y=sqrt(-x^2+6x)-2,y=-sqrt(-x^2+6x)-2
Find the zeros of solvefor y,(x-3)^2+(y+2)^2=9
The zeros of solvefor y,(x-3)^2+(y+2)^2=9 are y=sqrt(-x^2+6x)-2,y=-sqrt(-x^2+6x)-2