Chapter EP, Problem 10.6.6EP

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 10.6.6EP

The standard form of equation is ${\left(x-5\right)}^2+{\left(y-1\right)}^2=7^2$ , the equation is Circle and graph is shown in figure (1).

Explanation of Solution

Given:

The equation is $x^2+y^2=10x+2y+23$ .

Calculation:

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

$\begin{array}{c}{x}^2+y^2=10x+2y+23\\ x^2+y^2-10x-2y-23=0\\ x^2-10x+y^2-2y-23=0\\ x^2-10x+25-25+y^2-2y+1-1-23=0\end{array}$

Simplify further.

$\begin{array}{c}{\left(x-5\right)}^2+{\left(y-1\right)}^2-49=0\\ {\left(x-5\right)}^2+{\left(y-1\right)}^2=49\\ {\left(x-5\right)}^2+{\left(y-1\right)}^2=7^2\end{array}$

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 $x^2+y^2=10x+2y+23$ or ${\left(x-5\right)}^2+{\left(y-1\right)}^2=7^2$ is as follows.

Figure (1)

Therefore, the standard form of equation is ${\left(x-5\right)}^2+{\left(y-1\right)}^2=7^2$ , the equation is Circle and graph is shown in figure (1).