SOLUTION: uh Hey, I was trying to find a circle tangent to the parabola y^2=4x with minimum radius but I keep getting stuck. Can you please help?

Your parabola has the symmetry axis horizontal and is opened to the right.

I will rotate the whole picture 90 degrees anti-clockwise and will consider
a traditional parabola opened up with vertical symmetry axis x= 0. 
So, my parabola is

    y = 4x^2.   (1)


Next, I will consider a circle of the radius "r" in the upper half-plane with 
the the center on y-axis x= 0 tangent to x-axis and inscribed into the parabola,
having only one common point (0,0) with the parabola.

My task is to find maximum possible value of "r".


For the parabola, I have equation (1).

For a circle, I have standard form equation

    x^2 + (y-r)^2 = r^2,    (2),

where "r" is the radius. Again, my task is to determine maximum possible "r" so
that two curves (1) and (2) gave only one common point (0,0).


Substitute y = 4x^2 from (1) into (2),  You will get

    x^2 + (4x^2-r)^2 = r^2.    (3)


Open parentheses in (3) and simplify step by step

    x^2 + 16x^4 - 8rx^2 + r^2 = r^2

    16x^4 - (8r-1)x^2 = 0,

    16x^4 = (8r-1)x^2.


Since you are looking for the solution x different from 0, you may cancel x^2 in both sides of the last equation.
You will get then

    16x^2 = (8r-1).   (4)


Now, left side (4) is always positive. Hence, equation (4) has a solution x only if 8r-1 is positive: 

    8r-1 > 0,  or  r > 1/8.


If r = 1/8  or  r <= 1/8,  equation (4) has no solution.


HENCE, r = 1/8 is the MAXIMUM possible radius of your circle.


At this point, the problem is just SOLVED in full.


ANWSER.  The maximum radius of the circle under given condition is r = 1/8.