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. \( x^{2}+y^{2}+6 x-8 y+34=0 \) 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 The radius of the circle is D. The graph is nonexistent.

To determine whether the equation \( x^{2}+y^{2}+6 x-8 y+34=0 \) represents a circle, we can rewrite it in the standard form of a circle's equation, which is \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) is the radius. First, we will rearrange the given equation: \[ x^{2} + 6x + y^{2} - 8y + 34 = 0 \] Next, we will complete the square for the \(x\) and \(y\) terms. 1. For the \(x\) terms: \(x^{2} + 6x\) - Take half of 6, which is 3, and square it to get 9. - Thus, \(x^{2} + 6x = (x + 3)^{2} - 9\). 2. For the \(y\) terms: \(y^{2} - 8y\) - Take half of -8, which is -4, and square it to get 16. - Thus, \(y^{2} - 8y = (y - 4)^{2} - 16\). Now substitute these back into the equation: \[ (x + 3)^{2} - 9 + (y - 4)^{2} - 16 + 34 = 0 \] Combine the constants: \[ (x + 3)^{2} + (y - 4)^{2} - 9 - 16 + 34 = 0 \] This simplifies to: \[ (x + 3)^{2} + (y - 4)^{2} + 9 = 0 \] Now, isolate the squared terms: \[ (x + 3)^{2} + (y - 4)^{2} = -9 \] Since the left side of the equation represents the sum of two squares, it is always non-negative. The right side is -9, which is negative. Therefore, there are no real solutions to this equation. Thus, the graph of the equation does not exist. The correct choice is: **D. The graph is nonexistent.**