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 } - 6 x - 6 y + 18 = 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 line.
b. the graph of the equation is a point.
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

Group \(x\) - terms and \(y\) - terms: \((x^{2}-6x)+(y^{2}-6y)+18 = 0\)
For the \(x\) - terms: \(x^{2}-6x=(x - 3)^{2}-9\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(b = 3\))
For the \(y\) - terms: \(y^{2}-6y=(y - 3)^{2}-9\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=y\), \(b = 3\))
Substitute back into the equation: \((x - 3)^{2}-9+(y - 3)^{2}-9+18=0\)

Step2: Simplify the equation

Simplify the left - hand side: \((x - 3)^{2}+(y - 3)^{2}-18 + 18=0\)
We get \((x - 3)^{2}+(y - 3)^{2}=0\)

Since the sum of two non - negative numbers \((x - 3)^{2}\geq0\) and \((y - 3)^{2}\geq0\) is zero if and only if \(x-3 = 0\) and \(y - 3=0\)

Answer:

B. The graph of the equation is a point.