Write the equation in the form \( (x-h)^{2}+(y-k)^{2}=c \). Then, if the equation repr radius. If the equation represents the degenerate case, give the solution set. Part: \( 0 / 2 \) Part 1 of 2 The equation in standard form is \( \square . ~ \) Next Part

To rewrite the equation in the form \( (x-h)^{2}+(y-k)^{2}=c \), we typically start with the general equation of a circle, which is \( x^2 + y^2 + Dx + Ey + F = 0 \). By completing the square for both \( x \) and \( y \), we can transform it into the desired form. When \( c > 0 \), the equation represents a circle with radius \( r = \sqrt{c} \). If \( c = 0 \), it indicates a point, and if \( c < 0 \), it stands for a degenerate case, meaning no real points satisfy the equation – essentially, the solution set is empty! So, it's crucial to determine the value of \( c \) to understand the shape represented by the equation. Feel free to provide the specific equation you're working with, and I can help you complete the calculation!