Find the Domain (x-6)^2+(y-5)^2=1

${(x - 6)}^{2} + {(y - 5)}^{2} = 1$

Step 1

Subtract ${(x - 6)}^{2}$ from both sides of the equation.

${(y - 5)}^{2} = 1 - {(x - 6)}^{2}$

Step 2

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

$y - 5 = \pm \sqrt{1 - {(x - 6)}^{2}}$

Step 3

Simplify $\pm \sqrt{1 - {(x - 6)}^{2}}$ .

Rewrite $1$ as $1^{2}$ .

$y - 5 = \pm \sqrt{1^{2} - {(x - 6)}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x - 6$ .

$y - 5 = \pm \sqrt{(1 + x - 6) (1 - (x - 6))}$

Simplify.

Subtract $6$ from $1$ .

$y - 5 = \pm \sqrt{(x - 5) (1 - (x - 6))}$

Apply the distributive property.

$y - 5 = \pm \sqrt{(x - 5) (1 - x - - 6)}$

Multiply $- 1$ by $- 6$ .

$y - 5 = \pm \sqrt{(x - 5) (1 - x + 6)}$

Add $1$ and $6$ .

$y - 5 = \pm \sqrt{(x - 5) (- x + 7)}$

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.

$y - 5 = \sqrt{(x - 5) (- x + 7)}$

Add $5$ to both sides of the equation.

$y = \sqrt{(x - 5) (- x + 7)} + 5$

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

$y - 5 = - \sqrt{(x - 5) (- x + 7)}$

Add $5$ to both sides of the equation.

$y = - \sqrt{(x - 5) (- x + 7)} + 5$

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

$y = \sqrt{(x - 5) (- x + 7)} + 5$

$y = - \sqrt{(x - 5) (- x + 7)} + 5$

$y = \sqrt{(x - 5) (- x + 7)} + 5$

$y = - \sqrt{(x - 5) (- x + 7)} + 5$

Step 5

Set the radicand in $\sqrt{(x - 5) (- x + 7)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 5) (- x + 7) \geq 0$

Step 6

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$- x + 7 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $- x + 7$ equal to $0$ and solve for $x$ .

Set $- x + 7$ equal to $0$ .

$- x + 7 = 0$

Solve $- x + 7 = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$- x = - 7$

Divide each term in $- x = - 7$ by $- 1$ and simplify.

Divide each term in $- x = - 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 7}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 7}{- 1}$

Simplify the right side.

Divide $- 7$ by $- 1$ .

$x = 7$

The final solution is all the values that make $(x - 5) (- x + 7) \geq 0$ true.

$x = 5, 7$

Use each root to create test intervals.

$x < 5$

$5 < x < 7$

$x > 7$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 5$ to see if it makes the inequality true.

Choose a value on the interval $x < 5$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$((0) - 5) (- (0) + 7) \geq 0$

The left side $- 35$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $5 < x < 7$ to see if it makes the inequality true.

Choose a value on the interval $5 < x < 7$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$((6) - 5) (- (6) + 7) \geq 0$

The left side $1$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 7$ to see if it makes the inequality true.

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$((10) - 5) (- (10) + 7) \geq 0$

The left side $- 15$ is less than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < 5$ False

$5 < x < 7$ True

$x > 7$ False

$x < 5$ False

$5 < x < 7$ True

$x > 7$ False

The solution consists of all of the true intervals.

$5 \leq x \leq 7$

Step 7

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[5, 7]$

Set-Builder Notation:

${x | 5 \leq x \leq 7}$

Step 8

Find the Domain (x-6)^2+(y-5)^2=1

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