Solve for y (z-4)^2+(y-6)^2=36

${(z - 4)}^{2} + {(y - 6)}^{2} = 36$

Step 1

Subtract ${(z - 4)}^{2}$ from both sides of the equation.

${(y - 6)}^{2} = 36 - {(z - 4)}^{2}$

Step 2

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

$y - 6 = \pm \sqrt{36 - {(z - 4)}^{2}}$

Step 3

Simplify $\pm \sqrt{36 - {(z - 4)}^{2}}$ .

Rewrite $36$ as $6^{2}$ .

$y - 6 = \pm \sqrt{6^{2} - {(z - 4)}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 6$ and $b = z - 4$ .

$y - 6 = \pm \sqrt{(6 + z - 4) (6 - (z - 4))}$

Simplify.

Subtract $4$ from $6$ .

$y - 6 = \pm \sqrt{(z + 2) (6 - (z - 4))}$

Apply the distributive property.

$y - 6 = \pm \sqrt{(z + 2) (6 - z - - 4)}$

Multiply $- 1$ by $- 4$ .

$y - 6 = \pm \sqrt{(z + 2) (6 - z + 4)}$

Add $6$ and $4$ .

$y - 6 = \pm \sqrt{(z + 2) (- z + 10)}$

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.

$y - 6 = \sqrt{(z + 2) (- z + 10)}$

Add $6$ to both sides of the equation.

$y = \sqrt{(z + 2) (- z + 10)} + 6$

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

$y - 6 = - \sqrt{(z + 2) (- z + 10)}$

Add $6$ to both sides of the equation.

$y = - \sqrt{(z + 2) (- z + 10)} + 6$

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

$y = \sqrt{(z + 2) (- z + 10)} + 6$

$y = - \sqrt{(z + 2) (- z + 10)} + 6$

$y = \sqrt{(z + 2) (- z + 10)} + 6$

$y = - \sqrt{(z + 2) (- z + 10)} + 6$

Solve for y (z-4)^2+(y-6)^2=36

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