Chapter 9.1, Problem 39E

To determine

To find:The x - and y -intercepts of the graph of the circle $x^2-2x+y^2-6y-27=0$ .

Answer to Problem 39E

The x -intercepts are $\left(1+2\sqrt{7},0\right)$ and $\left(1-2\sqrt{7},0\right)$ , and the y -intercepts are $\left(0,-3\right)$ and $\left(0,9\right)$ .

Explanation of Solution

Given information:

The equation of the circle is $x^2-2x+y^2-6y-27=0$ .

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.

  $\begin{array}{c}{x}^2-2x+y^2-6y-27=0\\ x^2-2x+1+y^2-6y+9=1+9+27\\ {\left(x-1\right)}^2+{\left(y-3\right)}^2=37\end{array}$

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

  $\begin{array}{c}{\left(x-1\right)}^2+{\left(0-3\right)}^2=37\\ {\left(x-1\right)}^2+9=37\\ {\left(x-1\right)}^2=28\\ x=1\pm 2\sqrt{7}\end{array}$

Therefore, the x -intercepts are $\left(1+2\sqrt{7},0\right)$ and $\left(1-2\sqrt{7},0\right)$ .

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

  $\begin{array}{c}{\left(0-1\right)}^2+{\left(y-3\right)}^2=37\\ {\left(y-3\right)}^2=37-1\\ {\left(y-3\right)}^2=36\\ \left(y-3\right)=\pm 6\\ y=-3,9\end{array}$

Therefore, the y -interceptsare $\left(0,-3\right)$ and $\left(0,9\right)$ .

Draw the graph for the circle.

  

From the graph, it is clear that the x -intercepts are $\left(1+2\sqrt{7},0\right)$ and $\left(1-2\sqrt{7},0\right)$ . The y -intercepts are $\left(0,-3\right)$ and $\left(0,9\right)$ .

Therefore, the x -intercepts are $\left(1+2\sqrt{7},0\right)$ and $\left(1-2\sqrt{7},0\right)$ , and the y -intercepts are $\left(0,-3\right)$ and $\left(0,9\right)$ .