Find the Domain (x-7)^2+(y-3)^2=1

${(x - 7)}^{2} + {(y - 3)}^{2} = 1$

Step 1

Subtract ${(x - 7)}^{2}$ from both sides of the equation.

${(y - 3)}^{2} = 1 - {(x - 7)}^{2}$

Step 2

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

$y - 3 = \pm \sqrt{1 - {(x - 7)}^{2}}$

Step 3

Simplify $\pm \sqrt{1 - {(x - 7)}^{2}}$ .

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

$y - 3 = \pm \sqrt{1^{2} - {(x - 7)}^{2}}$

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 $b = x - 7$ .

$y - 3 = \pm \sqrt{(1 + x - 7) (1 - (x - 7))}$

Simplify.

Subtract $7$ from $1$ .

$y - 3 = \pm \sqrt{(x - 6) (1 - (x - 7))}$

Apply the distributive property.

$y - 3 = \pm \sqrt{(x - 6) (1 - x - - 7)}$

Multiply $- 1$ by $- 7$ .

$y - 3 = \pm \sqrt{(x - 6) (1 - x + 7)}$

Add $1$ and $7$ .

$y - 3 = \pm \sqrt{(x - 6) (- x + 8)}$

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.

$y - 3 = \sqrt{(x - 6) (- x + 8)}$

Add $3$ to both sides of the equation.

$y = \sqrt{(x - 6) (- x + 8)} + 3$

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

$y - 3 = - \sqrt{(x - 6) (- x + 8)}$

Add $3$ to both sides of the equation.

$y = - \sqrt{(x - 6) (- x + 8)} + 3$

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

$y = \sqrt{(x - 6) (- x + 8)} + 3$

$y = - \sqrt{(x - 6) (- x + 8)} + 3$

$y = \sqrt{(x - 6) (- x + 8)} + 3$

$y = - \sqrt{(x - 6) (- x + 8)} + 3$

Step 5

Set the radicand in $\sqrt{(x - 6) (- x + 8)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 6) (- x + 8) \geq 0$

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 $0$ .

$x - 6 = 0$

$- x + 8 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $- x + 8$ equal to $0$ and solve for $x$ .

Set $- x + 8$ equal to $0$ .

$- x + 8 = 0$

Solve $- x + 8 = 0$ for $x$ .

Subtract $8$ from both sides of the equation.

$- x = - 8$

Divide each term in $- x = - 8$ by $- 1$ and simplify.

Divide each term in $- x = - 8$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 8}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 8}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 8}{- 1}$

Simplify the right side.

Divide $- 8$ by $- 1$ .

$x = 8$

The final solution is all the values that make $(x - 6) (- x + 8) \geq 0$ true.

$x = 6, 8$

Use each root to create test intervals.

$x < 6$

$6 < x < 8$

$x > 8$

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 $x < 6$ to see if it makes the inequality true.

Choose a value on the interval $x < 6$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) - 6) (- (0) + 8) \geq 0$

The left side $- 48$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $6 < x < 8$ to see if it makes the inequality true.

Choose a value on the interval $6 < x < 8$ and see if this value makes the original inequality true.

$x = 7$

Replace $x$ with $7$ in the original inequality.

$((7) - 6) (- (7) + 8) \geq 0$

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 $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

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

$((10) - 6) (- (10) + 8) \geq 0$

The left side $- 8$ 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.

$x < 6$ False

$6 < x < 8$ True

$x > 8$ False

$x < 6$ False

$6 < x < 8$ True

$x > 8$ False

The solution consists of all of the true intervals.

$6 \leq x \leq 8$

Step 7

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

Interval Notation:

$[6, 8]$

Set-Builder Notation:

${x | 6 \leq x \leq 8}$

Step 8

Find the Domain (x-7)^2+(y-3)^2=1

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