Write the equation in the form \( (x-h)^{2}+(y-k)^{2}=c \). Then, if the equation degenerate case, give the solution set. \[ x^{2}+y^{2}-22 x+6 y+129=0 \] Part: \( 0 / 2 \)

To rewrite the given equation \( x^{2}+y^{2}-22 x+6 y+129=0 \) in the desired form, we start by reorganizing it into a standard circle equation. 1. Rearranging the equation, we have: \[ x^{2} - 22x + y^{2} + 6y + 129 = 0 \] 2. Next, we complete the square for the \( x \) and \( y \) terms. - For \( x^{2} - 22x \): \[ x^{2} - 22x = (x - 11)^{2} - 121 \] - For \( y^{2} + 6y \): \[ y^{2} + 6y = (y + 3)^{2} - 9 \] 3. Substitute these into the equation: \[ (x - 11)^{2} - 121 + (y + 3)^{2} - 9 + 129 = 0 \] \[ (x - 11)^{2} + (y + 3)^{2} - 1 = 0 \] 4. Finally, we arrange it to the circle form: \[ (x - 11)^{2} + (y + 3)^{2} = 1 \] Now, let's examine if the equation represents a degenerate case: Since the radius \( r \) from the equation \( (x-h)^{2}+(y-k)^{2}=c \) is given by the square root of \( c \), here \( c = 1 \) which means the circle has a radius of \( 1 \). Thus, this is not a degenerate case (that would occur with \( c = 0 \)). The solution set describes a circle, specifically all points \( (x, y) \) such that they are \( 1 \) unit away from the center \( (11, -3) \): Hence, the set of solutions is: \[ \{ (x, y) \mid (x - 11)^{2} + (y + 3)^{2} = 1 \} \] which represents the circumference of the circle centered at \( (11, -3) \) with a radius of \( 1 \).