Maths problem: would you mind to teach me how to do (b) i and ii, thanks?

(b)(i)  I'll start with the result of (a) --
CD^2 = AD*BD =>
(x - 4)^2 = (6 - y)*(x), just used info given in the problem.

Now algebra:
x^2 - 8x + 16 = 6x - xy =>
x^2 - 14x + xy + 16 = 0 =>
x^2 + (y - 14)x + 16 = 0.   That's the end of part (b)(i).

(ii) I'm not sure what they mean by "even value."  Does that mean the number 4?  If so, then
x^2 - 10x + 16 = 0 => x = 8 (because it can't be 2).
Then the point (x,y) is (8,4).
Then CD = 4 and AD = 2, so by part(a) you have BD = 8.
Now you have 3 points on the circle:
(4,6), (8,4), (4,-4).
The circle does intersect the x axis.
Its radius is 5 and its center is at (4,1).
Its equation is (x-4)^2 + (y-1)^2 = 25.