Precalculus Examples

$x^{2} + y^{2} - 6 y - 8 = 0$

Step 1

Add $8$ to both sides of the equation.

$x^{2} + y^{2} - 6 y = 8$

Step 2

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

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Step 2.1

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 2.2

Consider the vertex form of a parabola.

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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Step 2.3.2.1

Factor $2$ out of $- 6$ .

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

Step 2.3.2.2

Cancel the common factors.

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Step 2.3.2.2.1

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

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

Step 2.3.2.2.2

Cancel the common factor.

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

Step 2.3.2.2.3

Rewrite the expression.

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

Step 2.3.2.2.4

Divide $- 3$ by $1$ .

$d = - 3$

Step 2.4

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

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Step 2.4.1

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 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

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

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

Step 2.4.2.1.2

Multiply $4$ by $1$ .

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

Step 2.4.2.1.3

Divide $36$ by $4$ .

$e = 0 - 1 \cdot 9$

Step 2.4.2.1.4

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Step 2.4.2.2

Subtract $9$ from $0$ .

$e = - 9$

Step 2.5

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

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

Step 3

Substitute ${(y - 3)}^{2} - 9$ for $y^{2} - 6 y$ in the equation $x^{2} + y^{2} - 6 y = 8$ .

$x^{2} + {(y - 3)}^{2} - 9 = 8$

Step 4

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

$x^{2} + {(y - 3)}^{2} = 8 + 9$

Step 5

Add $8$ and $9$ .

$x^{2} + {(y - 3)}^{2} = 17$

Step 6

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 7

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.

$r = \sqrt{17}$

$h = 0$

$k = 3$

Step 8

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

Center: $(0, 3)$

Step 9

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

Center: $(0, 3)$

Radius: $\sqrt{17}$

Step 10