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. \( x^{2}+y^{2}-6 x-8 y+25=0 \) A. The graph of the equation is a line. B. The graph of the equation is a circle with center A. The radius of the circle is C. The graph of the equation is a point. D. The graph is nonexistent.

To determine whether the given equation 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 of the circle and \(r\) is the radius. The given equation is: \[ x^{2} + y^{2} - 6x - 8y + 25 = 0 \] First, we will rearrange the equation: \[ x^{2} - 6x + y^{2} - 8y + 25 = 0 \] Next, we will complete the square for the \(x\) and \(y\) terms. 1. For \(x^{2} - 6x\): - Take half of \(-6\) (which is \(-3\)), square it to get \(9\). - Thus, \(x^{2} - 6x\) can be rewritten as \((x - 3)^{2} - 9\). 2. For \(y^{2} - 8y\): - Take half of \(-8\) (which is \(-4\)), square it to get \(16\). - Thus, \(y^{2} - 8y\) can be rewritten as \((y - 4)^{2} - 16\). Now substituting these back into the equation: \[ (x - 3)^{2} - 9 + (y - 4)^{2} - 16 + 25 = 0 \] Simplifying this: \[ (x - 3)^{2} + (y - 4)^{2} - 9 - 16 + 25 = 0 \] \[ (x - 3)^{2} + (y - 4)^{2} + 0 = 0 \] This simplifies to: \[ (x - 3)^{2} + (y - 4)^{2} = 0 \] The equation \((x - 3)^{2} + (y - 4)^{2} = 0\) represents a circle with a radius of \(0\). This means that the graph is actually a single point. Thus, the center of the circle is \((3, 4)\) and the radius is \(0\). The correct answer is: C. The graph of the equation is a point. The center is \((3, 4)\) and the radius is \(0\).