Chapter 9.2, Problem 20WE

To determine

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

Answer to Problem 20WE

The required center and radius of the circle is $\left(-5,2\right)$ and $3$ respectively.

Explanation of Solution

Given information: The given equation of the circle is $x^2+y^2+10x-4y+20=0$ .

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.

  $\begin{array}{l}{x}^2+y^2+10x-4y+20=0\\ x^2+10x+y^2-4y+20=0\end{array}$

Make them perfect square,

  $\begin{array}{l}{x}^2+10x+20+5+y^2-4y+4-9=0\\ \left(x^2+10x+25\right)+\left(y^2-4y+4\right)=9\\ {\left(x+5\right)}^2+{\left(y-2\right)}^2=3^2\end{array}$

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 ${\left(x+5\right)}^2+{\left(y-2\right)}^2=3^2$ to ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Hence,

  $\begin{array}{l}\Rightarrow (h,k)=\left(-5,2\right)\\ \Rightarrow r^2=3^2\\ \Rightarrow r=3\end{array}$

Therefore, the required center and radius of the circle is $\left(-5,2\right)$ and $3$ respectively.