Chapter 1, Problem 9CT

To determine

The center and radius of the circle $x^2+y^2+4x-2y-4=0$, and graph the circle.

Answer to Problem 9CT

Solution:

The center of the circle is $\left(-2,1\right)$, and its radius is 3.

The graph of the circle is:

Explanation of Solution

Given Information:

The equation of the circle $x^2+y^2+4x-2y-4=0$.

Explanation:

Consider $x^2+y^2+4x-2y-4=0$.

By grouping the terms involving $x$ and involving $y$, and taking the constant to the right side,

$\Rightarrow \left(x^2+4x\right)+\left(y^2-2y\right)=4$

Complete the square of each expression in parenthesis. For that, add ${\left(\frac{4}{2}\right)}^2={\left(2\right)}^2=4$ in parenthesis involving $x$ terms, and add ${\left(\frac{2}{2}\right)}^2={\left(1\right)}^2=1$ in parenthesis involving $y$ terms, and also add 4 and 1 to the left side.

$\Rightarrow \left(x^2+4x+4\right)+\left(y^2-2y+1\right)=4+4+1$

$\Rightarrow \left(x^2+4x+4\right)+\left(y^2-2y+1\right)=9$

$\Rightarrow {\left(x+2\right)}^2+{\left(y-1\right)}^2=3^2$

By comparing ${\left(x+2\right)}^2+{\left(y-1\right)}^2=3^2$ with standard form of the equation of circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, it gives $h=-2,k=1$ and $r=3$.

Therefore, the center of the circle is $\left(-2,1\right)$, and its radius is 3.

The four points on the graph are $\left(-2+3,1\right)\equiv (1,1),\left(-2-3,1\right)\equiv (-5,1),\left(-2,1+3\right)\equiv (-2,4)$ and

$\left(-2,1-3\right)\equiv (-2,-2)$.

Thus the graph of the circle is: