Solve for y x^2+(y-2)^2=4

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

Step 1

Subtract $x^{2}$ from both sides of the equation.

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

Step 2

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y - 2 = \pm \sqrt{4 - x^{2}}$

Step 3

Simplify $\pm \sqrt{4 - x^{2}}$ .

Rewrite $4$ as $2^{2}$ .

$y - 2 = \pm \sqrt{2^{2} - x^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$y - 2 = \pm \sqrt{(2 + x) (2 - x)}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y - 2 = \sqrt{(2 + x) (2 - x)}$

Add $2$ to both sides of the equation.

$y = \sqrt{(2 + x) (2 - x)} + 2$

Next, use the negative value of the $\pm$ to find the second solution.

$y - 2 = - \sqrt{(2 + x) (2 - x)}$

Add $2$ to both sides of the equation.

$y = - \sqrt{(2 + x) (2 - x)} + 2$

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{(2 + x) (2 - x)} + 2$

$y = - \sqrt{(2 + x) (2 - x)} + 2$

$y = \sqrt{(2 + x) (2 - x)} + 2$

$y = - \sqrt{(2 + x) (2 - x)} + 2$