Problem 52 - 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. $$x^{2}+y^{2}-4 x-12 y-9=0$$

Step-by-Step Solution

Answer

The equation in standard form is \((x - 2)^{2} + (y - 6)^{2} = 31\). The center (h,k) of the circle is (2,6) and the radius \(r\) is \(\sqrt{31}\).

Step 1: Reorder the Equation

First, rearrange the equation to group the \(x\) and \(y\) terms together: \(x^{2}-4x+y^{2}-12y=-9\).

Step 2: Complete the Square for each Variable

To complete the square, add the square of half the coefficient of each \(x\) and \(y\) to both sides. For \(x^{2}-4x\), add \(((-4)/2)^{2}\) which is \(4\) to both sides of the equation. For \(y^{2}-12y\), add \(((-12)/2)^{2}\) which is \(36\). So, the new equation becomes \(x^{2}-4x+4+y^{2}-12y+36= -9 + 4 + 36\).

Step 3: Rewrite as Squares

Rewrite the equation as squares: \((x-2)^{2} + (y-6)^{2}= 31\).

Step 4: Identify the Center and Radius

The center (h,k) is given by the terms inside the parentheses with the opposite signs, and the radius \(r\) is the square root of the constant term on the right side of the equation: \(r= \sqrt{31}\). So, the center is \((2,6)\) and the radius is \(\sqrt{31}\).

Step 5: Plot the Circle

To graph the circle: Place a point at the center coordinates (2,6), and draw a circle with a radius of \(\sqrt{31}\)