What are the solutions to the equation solvefor y,(x-2)^2+(y+3)^2=4 ?

The solutions to the equation solvefor y,(x-2)^2+(y+3)^2=4 are y=sqrt(-x^2+4x)-3,y=-sqrt(-x^2+4x)-3

Find the zeros of solvefor y,(x-2)^2+(y+3)^2=4

The zeros of solvefor y,(x-2)^2+(y+3)^2=4 are y=sqrt(-x^2+4x)-3,y=-sqrt(-x^2+4x)-3

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solve for

y, (x - 2)^{2} + (y + 3)^{2} = 4

y = - x^{2} + 4 x - 3, y = - - x^{2} + 4 x - 3

Solution steps

(x - 2)^{2} + (y + 3)^{2} = 4

Move

(x - 2)^{2}

to the right side

(y + 3)^{2} = 4 - (x - 2)^{2}

Simplify

4 - (x - 2)^{2} : - x^{2} + 4 x

For

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

the solutions are

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

Solve

y + 3 = - x^{2} + 4 x : y = - x^{2} + 4 x - 3

Solve

y + 3 = - - x^{2} + 4 x : y = - - x^{2} + 4 x - 3

The solutions to the quadratic equation are:

y = - x^{2} + 4 x - 3, y = - - x^{2} + 4 x - 3

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

7 x^{2} = 1

- 2 (x - 4)^{2} = 11

x^{2} - 11 x + 12 = 0

9 x^{2} + 12 x + 4 = 5

What are the solutions to the equation solvefor y,(x-2)^2+(y+3)^2=4 ?
The solutions to the equation solvefor y,(x-2)^2+(y+3)^2=4 are y=sqrt(-x^2+4x)-3,y=-sqrt(-x^2+4x)-3
Find the zeros of solvefor y,(x-2)^2+(y+3)^2=4
The zeros of solvefor y,(x-2)^2+(y+3)^2=4 are y=sqrt(-x^2+4x)-3,y=-sqrt(-x^2+4x)-3