Problem 56 - 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 circle's formula in standard form is \((x-2)^{2} +(y-6)^{2} = 49\), the center of the circle is at the point (2, 6), and the radius of the circle is 7.

Step 1: Identifying the x and y terms

From the given equation \(x^{2}+y^{2}-4 x-12 y-9=0\), identify the x's and y's terms which are \(x^{2}-4 x\) and \(y^{2}-12 y\) respectively.

Step 2: Applying the 'Completing the Square' method

Now sort them individually by using the Complete the Square. For the x's terms, add and subtract the square of half the x's coefficient within the brackets: \(x^{2}-4 x+\left(\frac{-4}{2}\right)^{2} -\left(\frac{-4}{2}\right)^{2} = (x-2)^{2} - 4\). Do the same for the y's terms: \(y^{2}-12 y+\left(\frac{-12}{2}\right)^{2} -\left(\frac{-12}{2}\right)^{2} = (y-6)^{2} - 36\)

Step 3: Re-writing all terms with the circle's standard form

Adding the results from the previous step to the remaining term from the original equation, the equation in standard form becomes: \((x-2)^{2} - 4 +(y-6)^{2} - 36 -9=0 \). Simplifying this gives us \((x-2)^{2} +(y-6)^{2} = 49\) which is the standard form equation for a circle where \((x-h)^2 + (y-k)^2 = r^2\).

Step 4: Extracting the circle's center and radius

Looking at the obtained standard form equation \((x-2)^{2} +(y-6)^{2} = 49\), we can read that the center (h, k) is (2, 6) and the radius r is \(sqrt{49} = 7\).

Step 5: Graph the circle

Plot the center point (2,6) on a graph. From there, draw a circle with a radius length of 7 units. It's sometimes helpful to mark a few key points 7 units in the different directions from the center and then draw a smooth circle through these points.