Evaluate (x-h)^2+(y-k)^2=r^2

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Subtract ${(y - k)}^{2}$ from both sides of the equation.

${(x - h)}^{2} = r^{2} - {(y - k)}^{2}$

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

$x - h = \pm \sqrt{r^{2} - {(y - k)}^{2}}$

Simplify $\pm \sqrt{r^{2} - {(y - k)}^{2}}$ .

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

$x - h = \pm \sqrt{(r + y - k) (r - (y - k))}$

Simplify.

Apply the distributive property.

$x - h = \pm \sqrt{(r + y - k) (r - y - - k)}$

Multiply $- - k$ .

Multiply $- 1$ by $- 1$ .

$x - h = \pm \sqrt{(r + y - k) (r - y + 1 k)}$

Multiply $k$ by $1$ .

$x - h = \pm \sqrt{(r + y - k) (r - y + k)}$

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.

$x - h = \sqrt{(r + y - k) (r - y + k)}$

Add $h$ to both sides of the equation.

$x = \sqrt{(r + y - k) (r - y + k)} + h$

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

$x - h = - \sqrt{(r + y - k) (r - y + k)}$

Add $h$ to both sides of the equation.

$x = - \sqrt{(r + y - k) (r - y + k)} + h$

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

$x = \sqrt{(r + y - k) (r - y + k)} + h$

$x = - \sqrt{(r + y - k) (r - y + k)} + h$

$x = \sqrt{(r + y - k) (r - y + k)} + h$

$x = - \sqrt{(r + y - k) (r - y + k)} + h$

Evaluate (x-h)^2+(y-k)^2=r^2

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