Chapter 9.1, Problem 40E

To determine

To find:The x - and y -intercepts of the graph of the circle $x^2+8x+y^2+2y+9=0$ .

Answer to Problem 40E

The x -intercepts are $\left(-4+\sqrt{7},0\right)$ and $\left(-4-\sqrt{7},0\right)$ , and there is no y -intercept exist.

Explanation of Solution

Given information:

The equation of the circle is $x^2+8x+y^2+2y+9=0$ .

Calculation:

x -intercept is defined as the point where the graph of an equation crosses the x -axis.

y -intercept is defined as the point where the graph of an equation crosses the y -axis.

Rearrange the equation.

  $\begin{array}{c}{x}^2+8x+y^2+2y+9=0\\ x^2+8x+16+y^2+2y+1=16+1-9\\ {\left(x+4\right)}^2+{\left(y+1\right)}^2=8\end{array}$

To calculate the x -intercept, substitute $y=0$ in the equation of the circle and solve.

  $\begin{array}{c}{\left(x+4\right)}^2+{\left(y+1\right)}^2=8\\ {\left(x+4\right)}^2+{\left(0+1\right)}^2=8\\ {\left(x+4\right)}^2=7\\ x=-4\pm \sqrt{7}\end{array}$

Therefore, the x -intercepts are $\left(-4+\sqrt{7},0\right)$ and $\left(-4-\sqrt{7},0\right)$ .

To calculate the y -intercept, substitute $x=0$ in the equation of the circle and solve.

  $\begin{array}{c}{\left(x+4\right)}^2+{\left(y+1\right)}^2=8\\ {\left(0+4\right)}^2+{\left(y+1\right)}^2=8\\ {\left(y+1\right)}^2=-8\end{array}$

Since the square of a real number cannot be negative. So, there is no y -intercept exist.

Draw the graph for the circle.

  

From the graph, it is clear that the x -intercepts are $\left(-4+\sqrt{7},0\right)$ and $\left(-4-\sqrt{7},0\right)$ , and there is no y -intercept exist.

Therefore, the x -intercepts are $\left(-4+\sqrt{7},0\right)$ and $\left(-4-\sqrt{7},0\right)$ , and there is no y -intercept exist.