Problem 53 - 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 standard form of the equation is \((x+4)^{2} + (y-1)^{2} = 25\). The center of the circle is (-4,1) and the radius is 5.

Step 1: Arrange terms

Rearrange the equation with x terms and y terms together: \(x^{2} + 8x + y^{2} - 2y = 8\).

Step 2: Complete the square for x terms

In the equation, complete the square for x terms by adding \((\frac{b}{2})^{2}\) on both sides, where b is the coefficient of x. This results in the equation: \(x^{2} + 8x + 16 = 8 + 16\). This simplifies to: \((x+4)^{2} = 24\).

Step 3: Complete the square for y terms

In the same way, complete the square for y terms by adding \((\frac{b}{2})^{2}\) on both sides, where b is the coefficient of y. This results in the equation: \(y^{2} - 2y + 1 = 24 + 1\). This simplifies to: \((y-1)^{2} = 25\).

Step 4: Write in standard form

Now add both equations: \((x+4)^{2} + (y-1)^{2} = 25\). This is the standard form of the equation.

Step 5: Identify center and radius

Based on the standard form of the equation, the center of the circle (h,k) is (-4,1) and the radius is \(r=\sqrt{25}=5\).

Step 6: Graph the equation

Draw graph on x-y plane with point (-4,1) as center and radius 5. The circle should touch points (1,1), (-4,6), (-4,-4), and (-9,1).