Chapter 4, Problem 10CR

To determine

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

Answer to Problem 10CR

Solution:

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

  

Explanation of Solution

Given information:

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

Here, the equation of the circle is $x^2+4x+y^2-2y-4=0$

  $\Rightarrow \left(x^2+4x\right)+\left(y^2-2y\right)-4=0$ .

Now, complete the square of each expression in the parentheses

The quadratic equation $a{x}^2+bx+c=0$ can make the complete square by adding ${\left(b/2a\right)}^2$ on both sides of the equation

For $\left(x^2+4x\right)$ , the value ${\left(b/2a\right)}^2={\left(4/2\right)}^2=4$

For $\left(y^2-2y\right)$ , the value ${\left(b/2a\right)}^2={\left(-2/2\right)}^2=1$

So, add $4+1$ on both sides of the equation of the circle

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

  $\Rightarrow {\left(x+2\right)}^2+{\left(y-1\right)}^2-4=5$

  $\Rightarrow {\left(x+2\right)}^2+{\left(y-1\right)}^2=5+4=9$

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

The standard form of the equation of the circle with center $\left(h,k\right)$ and radius $r$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Compare the given equation with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

  $\Rightarrow h=-2,k=1,r=3$

Hence, the center is $\left(-2,1\right)$ and the radius is $r=3$

To draw the graph of the circle,

First, plot the center $\left(-2,1\right)$

Here, the radius is $r=3$ , locate four points on the circle by plotting points $3$ units to the left, to the right, up, and down from the center.

Now, draw the circle passing through these points.

The graph of the circle is given as