Chapter 13.1, Problem 45WE

To determine

To find the center and radius of a circle

Answer to Problem 45WE

The value of center are $\left(-2,4\right)$ and radius are 6

Explanation of Solution

The given equation is

  $x^2+4x+y^2-8y=16$

Calculation:

The center and radius of a circle can be obtained as

  $x^2+4x+y^2-8y=16$

First we made the equation to the standard form of the equation

  $\begin{array}{l}{x}^2+4x+y^2-8y=16\\ x^2+4x+4+y^2-8y+16=16+20\left[\text{add both sides 20}\right]\\ \left[\left(x^2\right)+2\left(2\right)\left(x\right)+{\left(2\right)}^2\right]+\left[{\left(y\right)}^2-2\left(4\right)\left(y\right)+{\left(4\right)}^2\right]=36\left[\text{perfect square trinomial}\right]\\ {\left(x+2\right)}^2+{\left(y-4\right)}^2={\left(6\right)}^2\end{array}$

The standard form of the equation of a circle with center and radius is

  ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Where $\left(a,b\right)$ is the center and r is the radius

Comparing given equation to the standard form of the equation

  $a=-2,b=4\text{ and }r=6$

Hence,

The value of center are $\left(-2,4\right)$ and radius are 6