Answer:
( x + 13 )^2 + ( y + 6 )^2 = 1; Option A
Step-by-step explanation:
~ The question we have at hand is: x^2 + y^2 + 26x + 12y + 204 = 0 ~
Let us apply the circle equation ( x - a )^2 + ( y - b )^2 = r^2 ⇒ provided r is the radius, centered at point ( a, b )
1. First rewrite x^2 + y^2 + 26x + 12y + 204 = 0 in the standard form of circle equation: x^2 + y^2 + 26x + 12y + 204 = 0
2. Now move the loose number to the right side: x^2 + 26x + y^2 + 12y = -204
3. Let us now group variables: ( x^2 + 26x ) + ( y^2 + 12y ) = -204
4. Convert x to square form: ( x^2 + 26x + 169 ) + ( y^2 + 12y ) = -204 + 169
5. Convert to square form: ( x + 13 )^2 + ( y^2 + 12y ) = -204 + 169
6. Convert y to square form: ( x + 13 )^2 + ( y^2 + 12y + 36 ) = -204 + 169 + 36
7. Convert to square form, and simplify: ( x + 13 )^2 + ( y + 6 )^2 = 1
Answer: ( x + 13 )^2 + ( y + 6 )^2 = 1
