Find the Center and Radius x^2+y^2-12x+2y+33=0

$x^{2} + y^{2} - 12 x + 2 y + 33 = 0$

Step 1

Subtract $33$ from both sides of the equation.

$x^{2} + y^{2} - 12 x + 2 y = - 33$

Step 2

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

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

Step 3

Substitute ${(x - 6)}^{2} - 36$ for $x^{2} - 12 x$ in the equation $x^{2} + y^{2} - 12 x + 2 y = - 33$ .

${(x - 6)}^{2} - 36 + y^{2} + 2 y = - 33$

Step 4

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

${(x - 6)}^{2} + y^{2} + 2 y = - 33 + 36$

Step 5

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

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

$a = 1$

$b = 2$

$c = 0$

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$d = 1$

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Subtract $1$ from $0$ .

$e = - 1$

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

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

Step 6

Substitute ${(y + 1)}^{2} - 1$ for $y^{2} + 2 y$ in the equation $x^{2} + y^{2} - 12 x + 2 y = - 33$ .

${(x - 6)}^{2} + {(y + 1)}^{2} - 1 = - 33 + 36$

Step 7

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

${(x - 6)}^{2} + {(y + 1)}^{2} = - 33 + 36 + 1$

Step 8

Simplify $- 33 + 36 + 1$ .

Add $- 33$ and $36$ .

${(x - 6)}^{2} + {(y + 1)}^{2} = 3 + 1$

Add $3$ and $1$ .

${(x - 6)}^{2} + {(y + 1)}^{2} = 4$

Step 9

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 10

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.

$r = 2$

$h = 6$

$k = - 1$

Step 11

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

Center: $(6, - 1)$

Step 12

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

Center: $(6, - 1)$

Radius: $2$

Step 13

Find the Center and Radius x^2+y^2-12x+2y+33=0

Related Questions

Support LCMGCF.com by disabling your adblocker.