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 - 8 y + 29 = 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: Group \(x\) and \(y\) terms

$$(x^{2}+6x)+(y^{2}-8y)+29 = 0$$

Step2: Complete the square for \(x\) terms

For \(x^{2}+6x\), we add \((\frac{6}{2})^{2}=9\).
For \(y^{2}-8y\), we add \((\frac{-8}{2})^{2}=16\).
Add \(9 + 16\) to both sides of the equation:
$$(x^{2}+6x + 9)+(y^{2}-8y+16)+29=9 + 16$$

Step3: Write in standard form

Using \((a + b)^{2}=a^{2}+2ab + b^{2}\) and \((a - b)^{2}=a^{2}-2ab + b^{2}\), we get:
$$(x + 3)^{2}+(y - 4)^{2}+29=25$$
$$(x + 3)^{2}+(y - 4)^{2}=- 4$$

Since the right - hand side of the equation \((x + 3)^{2}+(y - 4)^{2}=-4\) is negative (and \((x + 3)^{2}\geq0\), \((y - 4)^{2}\geq0\) for all real \(x\) and \(y\)), there are no real values of \(x\) and \(y\) that satisfy the equation.

Answer:

D. The graph is nonexistent.