Final answer:
The equation x²-6x+y²-12y-55 can be solved by completing the square, resulting in the equation of a circle, (x-3)² + (y-6)² = 100, with a radius of 10 centered at (3,6).
Step-by-step explanation:
The problem given is to solve the equation by completing the square for the quadratic terms of x and y. The original equation is x²-6x+y²-12y-55. To complete the square for the x-terms, we find the value that makes x²-6x a perfect square trinomial. This value is (-6/2)² = 9, so we add and subtract 9 to get x²-6x+9, which can be written as (x-3)².
Similarly, for the y-terms, we have y²-12y, and we need to add and subtract (-12/2)² = 36 to complete the square, resulting in y²-12y+36 which simplifies to (y-6)².
Substituting these expressions back into the original equation, we get (x-3)² + (y-6)² - 9 - 36 - 55 = 0, simplifying to (x-3)² + (y-6)² = 100. This represents the equation of a circle with a radius of 10 centered at (3,6). Therefore, the equation x²-6x+y²-12y-55 after completing the square becomes the equation of a circle: (x-3)² + (y-6)² = 100.