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. $$x^{2}+y^{2}+12 x-6 y-4=0$$

Step-by-Step Solution

Answer

The equation \(x^{2} + y^{2} + 12x -6y -4 = 0\) transforms to the standard form \((x+6)^{2} + (y-3)^{2} = 49\). The center of the circle is (-6, 3) and the radius is 7.

Step 1: Group and Rearrange Terms

Group and rearrange terms by \(x\) and \(y\). The equation \(x^{2} + 12x + y^{2} - 6y = 4\) is obtained.

Step 2: Complete the Square

Complete the square for both \(x\) and \(y\). Half of the coefficient of the \(x\) term is \((12/2)^{2} = 36\) and that for \(y\) term is \((-6/2)^{2} = 9\). Add these to both sides of the equation to balance it. The equation \( (x^{2} + 12x + 36) + (y^{2} - 6y + 9) = 4 + 36 + 9\) is obtained.

Step 3: Write the Equation in Standard Form

Write the equation in the form 'square plus square equals constant' to reflect the standard form of a circle equation. The standard form of the equation is \((x+6)^{2} + (y-3)^{2} = 49\).

Step 4: Identify the Center and Radius

The center and radius of the circle can be derived from the standard form of the equation. Compare with the general form \((x-a)^2 + (y-b)^2 = r^2\), the center \((a,b)\) is \((-6, 3)\) and the radius \(r\) is \(\sqrt{49} = 7\).

Step 5: Graph the Equation

Plot the center point at (-6,3) and draw a circle with radius of 7.