Find the Center x^2+y^2+8x+4y-3=40

$x^{2} + y^{2} + 8 x + 4 y - 3 = 40$

Step 1

Move all terms not containing a variable to the right side of the equation.

Add $3$ to both sides of the equation.

$x^{2} + y^{2} + 8 x + 4 y = 40 + 3$

Add $40$ and $3$ .

$x^{2} + y^{2} + 8 x + 4 y = 43$

Step 2

Complete the square for $x^{2} + 8 x$ .

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

$a = 1$

$b = 8$

$c = 0$

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 $d = \frac{b}{2 a}$ .

$d = \frac{8}{2 \cdot 1}$

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot 4}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{4}{1}$

Divide $4$ by $1$ .

$d = 4$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{8^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $8$ to the power of $2$ .

$e = 0 - \frac{64}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{64}{4}$

Divide $64$ by $4$ .

$e = 0 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 0 - 16$

Subtract $16$ from $0$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 4)}^{2} - 16$ .

${(x + 4)}^{2} - 16$

Step 3

Substitute ${(x + 4)}^{2} - 16$ for $x^{2} + 8 x$ in the equation $x^{2} + y^{2} + 8 x + 4 y = 43$ .

${(x + 4)}^{2} - 16 + y^{2} + 4 y = 43$

Step 4

Move $- 16$ to the right side of the equation by adding $16$ to both sides.

${(x + 4)}^{2} + y^{2} + 4 y = 43 + 16$

Step 5

Complete the square for $y^{2} + 4 y$ .

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

$a = 1$

$b = 4$

$c = 0$

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 $d = \frac{b}{2 a}$ .

$d = \frac{4}{2 \cdot 1}$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{2}{1}$

Divide $2$ by $1$ .

$d = 2$

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 $e = c - \frac{b^{2}}{4 a}$ .

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $4^{2}$ and $4$ .

Factor $4$ out of $4^{2}$ .

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

Cancel the common factors.

Factor $4$ out of $4 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $4$ by $1$ .

$e = 0 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 0 - 4$

Subtract $4$ from $0$ .

$e = - 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 2)}^{2} - 4$ .

${(y + 2)}^{2} - 4$

Step 6

Substitute ${(y + 2)}^{2} - 4$ for $y^{2} + 4 y$ in the equation $x^{2} + y^{2} + 8 x + 4 y = 43$ .

${(x + 4)}^{2} + {(y + 2)}^{2} - 4 = 43 + 16$

Step 7

Move $- 4$ to the right side of the equation by adding $4$ to both sides.

${(x + 4)}^{2} + {(y + 2)}^{2} = 43 + 16 + 4$

Step 8

Simplify $43 + 16 + 4$ .

Add $43$ and $16$ .

${(x + 4)}^{2} + {(y + 2)}^{2} = 59 + 4$

Add $59$ and $4$ .

${(x + 4)}^{2} + {(y + 2)}^{2} = 63$

Step 9

Divide each term by $63$ to make the right side equal to one.

$\frac{{(x + 4)}^{2}}{63} + \frac{{(y + 2)}^{2}}{63} = \frac{63}{63}$

Step 10

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 $1$ .

$\frac{{(x + 4)}^{2}}{63} + \frac{{(y + 2)}^{2}}{63} = 1$

Find the Center x^2+y^2+8x+4y-3=40

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