Question

decide whether or not the equation has a circle as its graph. if it does, give the center and the radius. if it does not, describe the graph.

$x^{2}+y^{2}-4x - 6y+13 = 0$

select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

a. the graph of the equation is a point.
b. the graph of the equation is a line.
c. the graph of the equation is a circle with center. (type an ordered pair.) the radius of the circle is.
d. the graph is nonexistent.

Explanation:

Step1: Complete the square for x and y terms.

\[

$$\begin{align*} x^{2}+y^{2}-4x - 6y+13&=0\\ x^{2}-4x + y^{2}-6y&=- 13\\ x^{2}-4x + 4+y^{2}-6y+9&=-13 + 4+9 \end{align*}$$

\]

Step2: Rewrite in standard - form.

\((x - 2)^{2}+(y - 3)^{2}=0\)

Answer:

A. The graph of the equation is a point.