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

Question

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!

Answer 1

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 }.$

Answer 2

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 ....

Answer 3

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 $$R=\dfrac{2+a^2}{\sqrt{4a^2+1}}=\dfrac12\sqrt{7}$$