Problem 54 - College Algebra

Question

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+8 x+4 y+16=0$$

Step-by-Step Solution

Answer

The center of the circle is at (-4,-2) and the radius is 2 units.

Step 1: Grouping the x's and y's

Separate and group the x terms and the y terms like this \(x^{2}+8x + y^{2}+4y=-16.\)

Step 2: Completing the square for x and y

Complete the square for x terms and y terms by halfing the coefficients of x and y, squaring it and adding it to both sides. To complete the square, we want to add \(((8/2)^2)=16\) to the x terms and \(((4/2)^2)=4\) to the y terms, like this: \(x^{2}+8x+16 + y^{2}+4y+4= -16+16+4.\)

Step 3: Simplifying the equation

After completing the square, we simplify the equation to:\((x+4)^{2}+ (y+2)^2= 4.\)

Step 4: Identifying the center and the radius

In the form \((x-h)^{2}+ (y-k)^2= r^2\), the center is located at (h,k), and the radius is \(r\). From our simplified equation, one can see that (h,k) is (-4,-2) and \(r=\sqrt{4}=2\) .

Step 5: Graphing the circle

To graph the circle, make a point at (-4,-2) for the center of the circle and draw the surrounding circle having a radius of 2 units.