Find the Center x^2-y^2-12x-10y-25=0

$x^{2} - y^{2} - 12 x - 10 y - 25 = 0$

Find the standard form of the hyperbola.

Add $25$ to both sides of the equation.

$x^{2} - y^{2} - 12 x - 10 y = 25$

Complete the square for $x^{2} - 12 x$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = - 12$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 12}{2 \cdot 1}$

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

$d = \frac{2 \cdot - 6}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot - 6}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot - 6}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 6}{1}$

Divide $- 6$ by $1$ .

$d = - 6$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 12)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $- 12$ to the power of $2$ .

$e = 0 - \frac{144}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{144}{4}$

Divide $144$ by $4$ .

$e = 0 - 1 \cdot 36$

Multiply $- 1$ by $36$ .

$e = 0 - 36$

Subtract $36$ from $0$ .

$e = - 36$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 6)}^{2} - 36$ .

${(x - 6)}^{2} - 36$

Substitute ${(x - 6)}^{2} - 36$ for $x^{2} - 12 x$ in the equation $x^{2} - y^{2} - 12 x - 10 y = 25$ .

${(x - 6)}^{2} - 36 - y^{2} - 10 y = 25$

Move $- 36$ to the right side of the equation by adding $36$ to both sides.

${(x - 6)}^{2} - y^{2} - 10 y = 25 + 36$

Complete the square for $- y^{2} - 10 y$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 1$

$b = - 10$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 10}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $- 10$ and $2$ .

Factor $2$ out of $- 10$ .

$d = \frac{2 \cdot - 5}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{- 5}{- 1}$ .

$d = - 1 \cdot - 5$

Rewrite $- 1 \cdot - 5$ as $- - 5$ .

$d = - - 5$

Multiply $- 1$ by $- 5$ .

$d = 5$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 10)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $- 10$ to the power of $2$ .

$e = 0 - \frac{100}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 0 - \frac{100}{- 4}$

Divide $100$ by $- 4$ .

$e = 0 - - 25$

Multiply $- 1$ by $- 25$ .

$e = 0 + 25$

Add $0$ and $25$ .

$e = 25$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y + 5)}^{2} + 25$ .

$- {(y + 5)}^{2} + 25$

Substitute $- {(y + 5)}^{2} + 25$ for $- y^{2} - 10 y$ in the equation $x^{2} - y^{2} - 12 x - 10 y = 25$ .

${(x - 6)}^{2} - {(y + 5)}^{2} + 25 = 25 + 36$

Move $25$ to the right side of the equation by adding $25$ to both sides.

${(x - 6)}^{2} - {(y + 5)}^{2} = 25 + 36 - 25$

Simplify $25 + 36 - 25$ .

Add $25$ and $36$ .

${(x - 6)}^{2} - {(y + 5)}^{2} = 61 - 25$

Subtract $25$ from $61$ .

${(x - 6)}^{2} - {(y + 5)}^{2} = 36$

Divide each term by $36$ to make the right side equal to one.

$\frac{{(x - 6)}^{2}}{36} - \frac{{(y + 5)}^{2}}{36} = \frac{36}{36}$

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(x - 6)}^{2}}{36} - \frac{{(y + 5)}^{2}}{36} = 1$

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 6$

$b = 6$

$k = - 5$

$h = 6$

The center of a hyperbola follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(6, - 5)$

Find the Center x^2-y^2-12x-10y-25=0

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