Chapter EP, Problem 10.6.3EP

To determine

To write: The equation in standard form, state the graph of the equation is a parabola, circle, ellipse, or hyperbola and then graph the equation.

Answer to Problem 10.6.3EP

The standard form of equation is ${\left(x+3\right)}^2+{\left(y-3\right)}^2={\left(3\right)}^2$ , the equation is Ellipse and graph is shown in figure (1).

Explanation of Solution

Given:

The equation is $x^2+6x+y^2-6y+9=0$ .

Calculation:

Rearrange the equation in the form of perfect square as follows.

$\begin{array}{c}{x}^2+6x+y^2-6y+9=0\\ x^2+2\left(3\right)x+9-9+y^2-2\left(3\right)y+9-9+9=0\\ {\left(x+3\right)}^2+{\left(y-3\right)}^2-9=0\\ {\left(x+3\right)}^2+{\left(y-3\right)}^2={\left(3\right)}^2\end{array}$

The general form of circle equation is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Therefore, the given equation is Circle.

The graph of the equation $x^2+6x+y^2-6y+9=0$ is as follows.

Figure (1)

Therefore, the standard form of equation is ${\left(x+3\right)}^2+{\left(y-3\right)}^2={\left(3\right)}^2$ , the equation is Ellipse and graph is shown in figure (1).