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. $$x^{2}+y^{2}-6 y-7=0$$

Step-by-Step Solution

Answer

The standard form of the given equation is \(x^{2}+(y-3)^{2}=16\). The center of the circle is at (0, 3) and the radius is 4 units.

Step 1: Regroup the equation

Rewrite the given equation \(x^{2}+y^{2}-6 y-7=0\) by regrouping the terms to get \(x^{2}+(y^{2}-6 y)-7=0\).

Step 2: Complete the square on the y terms

To complete the square on \(y^{2}-6 y\), we need to take half of the coefficient of y (-6 in this case), square it and add to both sides. Half of -6 is -3 and square of -3 is 9. Our equation now becomes \(x^{2}+(y^{2}-6y+9) = 7+9\) which simplifies to \(x^{2}+(y-3 )^{2} = 16\).

Step 3: Identify the Center and Radius

From the standard form \(x^{2}+(y-3)^{2}=16\), we can see that the center of the circle is at (0, 3) and the radius is \(sqrt{16}\) which equals to 4.

Step 4: Graph the Circle

Draw a graph with the x and y-axis. Mark the center at (0, 3). From the center, draw a circle with a radius of 4 units in all directions.