A circle centered at $(0,2)$ is tangent to $y=x^2$ at exactly two points. What is its radius? 
A circle is centered at $(0,2)$ and is tangent to $y=x^2$ at exactly two points. What is the radius of the circle? 

Don't really have an idea at how to solve the problem. Help is appreciated! 
 A: The circle has equation
$$x^2+(y-2)^2=r^2.$$
It meets the parabola at the points with $x$-coordinates satisfying
$$x^2+(x^2-2)^2=r^2.\tag1$$
Although this is a quartic equation, it is quadratic in $x^2$.
Thus
$$(z-2)^2+z-r^2=0\tag2$$
for $z=x^2$
For the circle to be tangent at two points, (1) must have two pairs of repeated roots, and so (2) has a repeated root. There is a well-known
criterion for a quadratic to have a repeated root ....
A: Let $(a,b)$ is the meet point which lies on $y=x^2$ so $b=a^2$. The tangent line on $y=x^2$ is of the form 
$$y-b=2a(x-a)$$
which is perpendicular to radius of circle, so the line passes through $(0,2)$ and $(a,b)$ is 
$$y=-\dfrac{x}{2a}+2$$
the intersection of these lines shows $a=\pm\sqrt{\dfrac32}$ then $R$ is the distance $(0,2)$ from tangent line $y=2ax-a^2$, that is 
$$R=\dfrac{2+a^2}{\sqrt{4a^2+1}}=\dfrac12\sqrt{7}$$
A: Let $(x,x^2)$ be a point on the parabola.
Line through $(0,2)$ and $(x,x^2)$ has slope:
$m:=\dfrac{2-x^2}{-x}.$
Slope of parabola at $x$: $m':= dy/dx =2x.$
$mm' =-1$, perpendicular.
$\dfrac{2-x^2}{-x} = -\dfrac{1}{2x}.$
$x=^+_-\sqrt{3/2},$ and  $y= 3/2$. $        $
(On the parabola);
Insert $(x,y)$ into
$x^2 +(y-2)^2=r^2 $
to find:
$ r= \sqrt{ 7/4 }.$
