The answer is zero sorry if I am wrong
If you can rewrite the formula as (x-a)² + (y-b)² = r², the center is at (a,b) and the radius is r.
If you work out this equation, and map it to the original, you will find that the +4x term hints that a = 2 (double product) and -12y hints that b=-6, and r=6.
So, the formula can be written as (x+2)² + (y-6)² = 6² and the center is at (-2,6) and the radius is 6.
What is the question here that I should answer?
Given :


Now, Substituing the value of y in equation (ii) :








Now, substituting the value of x in equation (i) :




Answer:
A. Use inverse relationships to "undo" the operation.
Step-by-step explanation:
I took the , and this was the correct answer. :)