Precalculus Examples

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

Step 1

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

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Step 1.1

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

$a = 1$

$b = 2$

$c = 0$

Step 1.2

Consider the vertex form of a parabola.

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

Cancel the common factor of $2$ .

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Step 1.3.2.1

Cancel the common factor.

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

Step 1.3.2.2

Rewrite the expression.

$d = 1$

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

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

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

Step 1.4.2.1.2

Multiply $4$ by $1$ .

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

Step 1.4.2.1.3

Cancel the common factor of $4$ .

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Step 1.4.2.1.3.1

Cancel the common factor.

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

Step 1.4.2.1.3.2

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Step 1.4.2.1.4

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Step 1.4.2.2

Subtract $1$ from $0$ .

$e = - 1$

Step 1.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 1)}^{2} - 1$ .

${(x + 1)}^{2} - 1$

Step 2

Substitute ${(x + 1)}^{2} - 1$ for $x^{2} + 2 x$ in the equation $x^{2} + y^{2} + 2 x + 6 y = - 1$ .

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

Step 3

Move $- 1$ to the right side of the equation by adding $1$ to both sides.

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

Step 4

Complete the square for $y^{2} + 6 y$ .

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Step 4.1

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

$a = 1$

$b = 6$

$c = 0$

Step 4.2

Consider the vertex form of a parabola.

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Cancel the common factor of $6$ and $2$ .

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Step 4.3.2.1

Factor $2$ out of $6$ .

$d = \frac{2 \cdot 3}{2 \cdot 1}$

Step 4.3.2.2

Cancel the common factors.

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Step 4.3.2.2.1

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

$d = \frac{2 \cdot 3}{2 (1)}$

Step 4.3.2.2.2

Cancel the common factor.

$d = \frac{2 \cdot 3}{2 \cdot 1}$

Step 4.3.2.2.3

Rewrite the expression.

$d = \frac{3}{1}$

Step 4.3.2.2.4

Divide $3$ by $1$ .

$d = 3$

Step 4.4

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

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Step 4.4.1

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

$e = 0 - \frac{6^{2}}{4 \cdot 1}$

Step 4.4.2

Simplify the right side.

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Step 4.4.2.1

Simplify each term.

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Step 4.4.2.1.1

Raise $6$ to the power of $2$ .

$e = 0 - \frac{36}{4 \cdot 1}$

Step 4.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{36}{4}$

Step 4.4.2.1.3

Divide $36$ by $4$ .

$e = 0 - 1 \cdot 9$

Step 4.4.2.1.4

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Step 4.4.2.2

Subtract $9$ from $0$ .

$e = - 9$

Step 4.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 3)}^{2} - 9$ .

${(y + 3)}^{2} - 9$

Step 5

Substitute ${(y + 3)}^{2} - 9$ for $y^{2} + 6 y$ in the equation $x^{2} + y^{2} + 2 x + 6 y = - 1$ .

${(x + 1)}^{2} + {(y + 3)}^{2} - 9 = - 1 + 1$

Step 6

Move $- 9$ to the right side of the equation by adding $9$ to both sides.

${(x + 1)}^{2} + {(y + 3)}^{2} = - 1 + 1 + 9$

Step 7

Simplify $- 1 + 1 + 9$ .

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Step 7.1

Add $- 1$ and $1$ .

${(x + 1)}^{2} + {(y + 3)}^{2} = 0 + 9$

Step 7.2

Add $0$ and $9$ .

${(x + 1)}^{2} + {(y + 3)}^{2} = 9$

Step 8

This is the form of a circle. Use this form to determine the center and radius of the circle.

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Step 9

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

$r = 3$

$h = - 1$

$k = - 3$

Step 10

The center of the circle is found at $(h, k)$ .

Center: $(- 1, - 3)$

Step 11

These values represent the important values for graphing and analyzing a circle.

Center: $(- 1, - 3)$

Radius: $3$

Step 12