Problem 80 - College Algebra Essentials

Question

Does \((x-3)^{2}+(y-5)^{2}=0\) represent the equation of a circle? If not, describe the graph of this equation.

Step-by-Step Solution

Answer

No, \((x-3)^{2}+(y-5)^{2}=0\) does not represent the equation of a circle with a nonzero radius. The equation represents a point located at (3,5) in the coordinate plane.

Step 1: Identify the form of the equation

The given equation \((x-3)^{2}+(y-5)^{2}=0\) is of the form \((x-h)^{2}+(y-k)^{2}=r^{2}\). Here, \(h=3\), \(k=5\), and \(r^{2}=0\). This is the standard form of the equation of a circle where \((h, k)\) is the center and \(r\) is the radius.

Step 2: Determine the values of h, k, and r

From the standard form, \(h=3\), \(k=5\) and \(r^{2}=0\). Taking the square root of \(r^{2}\), we find that \(r=0\).

Step 3: Describe the graph

Since the radius of the circle is 0, the circle degenerates to a single point. Therefore, the graph of the equation will be a single point at the location \((h,k)\) or in this case, \((3,5)\).