Precalculus Examples

$4 x^{2} + 4 y^{2} + 4 x + 4 y - 2 = 0$

Step 1

Add $2$ to both sides of the equation.

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

Step 2

Divide both sides of the equation by $4$ .

$x^{2} + y^{2} + x + y = \frac{1}{2}$

Step 3

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

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Step 3.1

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

$a = 1$

$b = 1$

$c = 0$

Step 3.2

Consider the vertex form of a parabola.

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Cancel the common factor of $1$ .

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Step 3.3.2.1

Cancel the common factor.

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

Step 3.3.2.2

Rewrite the expression.

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

One to any power is one.

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

Step 3.4.2.1.2

Multiply $4$ by $1$ .

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

Step 3.4.2.2

Subtract $\frac{1}{4}$ from $0$ .

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

Step 3.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + \frac{1}{2})}^{2} - \frac{1}{4}$ .

${(x + \frac{1}{2})}^{2} - \frac{1}{4}$

Step 4

Substitute ${(x + \frac{1}{2})}^{2} - \frac{1}{4}$ for $x^{2} + x$ in the equation $x^{2} + y^{2} + x + y = \frac{1}{2}$ .

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

Step 5

Move $- \frac{1}{4}$ to the right side of the equation by adding $\frac{1}{4}$ to both sides.

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

Step 6

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

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Step 6.1

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

$a = 1$

$b = 1$

$c = 0$

Step 6.2

Consider the vertex form of a parabola.

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

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Cancel the common factor of $1$ .

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Step 6.3.2.1

Cancel the common factor.

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

Step 6.3.2.2

Rewrite the expression.

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

Step 6.4

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

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Step 6.4.1

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

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

Step 6.4.2

Simplify the right side.

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Step 6.4.2.1

Simplify each term.

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Step 6.4.2.1.1

One to any power is one.

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

Step 6.4.2.1.2

Multiply $4$ by $1$ .

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

Step 6.4.2.2

Subtract $\frac{1}{4}$ from $0$ .

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

Step 6.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + \frac{1}{2})}^{2} - \frac{1}{4}$ .

${(y + \frac{1}{2})}^{2} - \frac{1}{4}$

Step 7

Substitute ${(y + \frac{1}{2})}^{2} - \frac{1}{4}$ for $y^{2} + y$ in the equation $x^{2} + y^{2} + x + y = \frac{1}{2}$ .

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

Step 8

Move $- \frac{1}{4}$ to the right side of the equation by adding $\frac{1}{4}$ to both sides.

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

Step 9

Simplify $\frac{1}{2} + \frac{1}{4} + \frac{1}{4}$ .

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Step 9.1

Combine the numerators over the common denominator.

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

Step 9.2

Add $1$ and $1$ .

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

Step 9.3

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

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Step 9.3.1

Factor $2$ out of $2$ .

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

Step 9.3.2

Cancel the common factors.

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Step 9.3.2.1

Factor $2$ out of $4$ .

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

Step 9.3.2.2

Cancel the common factor.

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

Step 9.3.2.3

Rewrite the expression.

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

Step 9.4

Combine the numerators over the common denominator.

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

Step 9.5

Simplify the expression.

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Step 9.5.1

Add $1$ and $1$ .

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

Step 9.5.2

Divide $2$ by $2$ .

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

Step 10

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 11

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.

$r = 1$

$h = - \frac{1}{2}$

$k = - \frac{1}{2}$

Step 12

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

Center: $(- \frac{1}{2}, - \frac{1}{2})$

Step 13

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

Center: $(- \frac{1}{2}, - \frac{1}{2})$

Radius: $1$

Step 14