Question 456905
The equation of the line is {{{y = x-1}}} rewritten. Hence a point on the line is (x, x-1). Let's say the center is at (x, x-1). Then the distance between (6,6) and the center = radius = distance between center and (0,-2). Using the point to point distance formula, we have 

{{{sqrt((x-6)^2 + (x-1-6)^2) = sqrt(x^2+(x-1+2)^2)}}}

This is the same as {{{(x-6)^2+(x-7)^2 = x^2 + (x+1)^2}}}

The solution is x = 3. Hence, the center is at (3,2).

The radius = 5 since {{{sqrt((6-3)^2 + (6-2)^2) = 5}}}.

So, the equation of the circle is {{{(x-3)^2 + (y-2)^2 = 5^2}}}