I did some work on Question 16 c
i) x^2 + (y-c)^2 = r^2
And y=x^2
Substitute 2 into 1
y + (y-c)^2= r^2
Rearrange (expand) to make a quadratic, y as the variable
y+y^2 -2yc + c^2 -r^2 = 0
Grouping terms
y^2 + y(1-2c) + (c^2-r^2) = 0
We know that there are two equal (in terms of y) solutions to this, because the circle and the parabola touch symmetrically around the y axis at the same value of y (given in the question). Therefore the discriminant of this quadratic = 0
discriminant = b^2-4ac
substituting in
(1-2c)^2 - 4(c^2-r^2) = 0
Expand and cancel (easy enough) to get 4c = 1 + 4r^2
In part ii) I got c>1/4, which is incorrect (we were meant to get c>1/2). Until I can figure this out I won't go any further.
Any further discussion of this question should go in the question 16 discussion thread, which I just discovered here:
http://community.boredofstudies.org/showthread.php?t=293638&page=4