Problem 57 - 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. $$x^{2}+y^{2}+8 x-2 y-8=0$$

Step-by-Step Solution

Answer

The equation in standard form is \( (x + 4)^{2} + (y - 1)^{2} = 25 \). Center of the circle is (-4, 1) and the radius is 5.

Step 1: Group the x and y terms together

First, let's group the x and y terms in the equation together. The equation \( x^{2}+y^{2}+8 x-2 y-8=0 \) becomes \( (x^{2} + 8x) + (y^{2} - 2y) = 8 \)

Step 2: Complete the square for x and y

Completing the square involves adding the square of half of the coefficient of x and y to both sides. This gives the equation \( (x^{2} + 8x + 16) + (y^{2} - 2y + 1) = 8 + 16 + 1 \), which simplifies to \( (x + 4)^{2} + (y - 1)^{2} = 25 \)

Step 3: Identify the center and radius of the circle

In the equation \( (x - h)^{2} + (y - k)^{2} = r^{2} \), (h, k) are the coordinates of the center of the circle, and r is the radius. From our equation \( (x + 4)^{2} + (y - 1)^{2} = 25 \), we can deduce that the center of the circle is (-4, 1) and the radius is 5

Step 4: Graph the equation

When graphing the equation, start by marking the center of the circle at the point (-4, 1). Then, draw a circle with radius 5 around this point.