Chapter 9.2, Problem 17WE

To determine

To calculate: The center and radius of the circle, if the given equation is a circle.

Answer to Problem 17WE

The required center and radius of the circle is $\left(0,-4\right)$ and $4$ respectively.

Explanation of Solution

Given information: The given equation of the circle is $x^2+y^2=-8y$ .

Formula used: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation: Rewrite the given equation into the standard form of a circle.

  $\begin{array}{l}{x}^2+y^2=-8y\\ x^2+y^2+8y=0\end{array}$

Make them perfect square,

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

Since it has the same coefficient of $x^2$ and $y^2$ terms. It is an equation of a circle.

Therefore, the given equation has a graph.

Now, we'll sketch the required graph of the circle.

  

As we know that the center of a circle is $(h,k)$ and radius is ‘ $r$ ’. Then, compare ${\left(x-0\right)}^2+\left(y+4\right)=4^2$ to ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Hence,

  $\begin{array}{l}\Rightarrow (h,k)=\left(0,-4\right)\\ \Rightarrow r^2=4^2\\ \Rightarrow r=4\end{array}$

Therefore, the required center and radius of the circle is $\left(0,-4\right)$ and $4$ respectively.