Chapter 9.1, Problem 36E

To determine

To find:The x - and y -intercepts of the graph of the circle ${\left(x-1\right)}^2+{\left(y+4\right)}^2=16$ .

Answer to Problem 36E

Thex -intercept is $\left(1,0\right)$ and the y -intercepts are $\left(0,-4-\sqrt{15}\right)$ and $\left(0,-4+\sqrt{15}\right)$ .

Explanation of Solution

Given information:

The equation of the circle is ${\left(x-1\right)}^2+{\left(y+4\right)}^2=16$ .

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.

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

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

Therefore, the x -interceptis $\left(1,0\right)$ .

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

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

Compare the equation with the general quadratic equation $a{x}^2+bx+c=0$ .

  $\begin{array}{l}a=1\\ b=8\\ c=1\end{array}$

Apply quadratic formula to calculate the values of $y$ .

  $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute the values in the above formula.

  $\begin{array}{c}y=\frac{-8\pm \sqrt{8^2-4\cdot 1\cdot 1}}{2\cdot 1}\\ =\frac{-8\pm \sqrt{64-4}}{2}\\ =\frac{-8\pm \sqrt{60}}{2}\\ =-4\pm \sqrt{15}\end{array}$

Therefore, the y -intercepts are $\left(0,-4-\sqrt{15}\right)$ and $\left(0,-4+\sqrt{15}\right)$ .

Draw the graph for the circle.

  

From the graph, it is clear that the x -intercept is $\left(1,0\right)$ , and the y -intercepts are $\left(0,-4-\sqrt{15}\right)$ and $\left(0,-4+\sqrt{15}\right)$ .

Therefore, the x -intercept is $\left(1,0\right)$ and the y -intercepts are $\left(0,-4-\sqrt{15}\right)$ and $\left(0,-4+\sqrt{15}\right)$ .