Find the Center and Radius x^2+y^2+4x-6y-3=0

$x^{2} + y^{2} + 4 x - 6 y - 3 = 0$

Step 1

Add $3$ to both sides of the equation.

$x^{2} + y^{2} + 4 x - 6 y = 3$

Step 2

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

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

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

Step 3

Substitute ${(x + 2)}^{2} - 4$ for $x^{2} + 4 x$ in the equation $x^{2} + y^{2} + 4 x - 6 y = 3$ .

${(x + 2)}^{2} - 4 + y^{2} - 6 y = 3$

Step 4

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

${(x + 2)}^{2} + y^{2} - 6 y = 3 + 4$

Step 5

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

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

$a = 1$

$b = - 6$

$c = 0$

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

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

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

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 3$ by $1$ .

$d = - 3$

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Divide $36$ by $4$ .

$e = 0 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Subtract $9$ from $0$ .

$e = - 9$

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

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

Step 6

Substitute ${(y - 3)}^{2} - 9$ for $y^{2} - 6 y$ in the equation $x^{2} + y^{2} + 4 x - 6 y = 3$ .

${(x + 2)}^{2} + {(y - 3)}^{2} - 9 = 3 + 4$

Step 7

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

${(x + 2)}^{2} + {(y - 3)}^{2} = 3 + 4 + 9$

Step 8

Simplify $3 + 4 + 9$ .

Add $3$ and $4$ .

${(x + 2)}^{2} + {(y - 3)}^{2} = 7 + 9$

Add $7$ and $9$ .

${(x + 2)}^{2} + {(y - 3)}^{2} = 16$

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.

$r = 4$

$h = - 2$

$k = 3$

Step 11

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

Center: $(- 2, 3)$

Step 12

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

Center: $(- 2, 3)$

Radius: $4$

Step 13

Find the Center and Radius x^2+y^2+4x-6y-3=0

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