<s>[INST] Largest equation of a circle that shares 2 tangents with a curve Just played around on a graphic calculator a little, and discovered that given the curve $y=x^2$ , all circles with the equations in the form of $\left(y-a\right)^2+x^2=\frac{4a-1}{4}$ for all $a>0.5$ share exactly 2 points with the curve i.e. it's the largest circle which 'fits' into the curve; here's a visualisation of what i mean: 
I'd like to know how you might go about finding this general formula without knowing the answer beforehand.
my proof:
substitute in $y$ for $x^2$
$\left(x^2-a\right)^2+x^2=\frac{4a-1}{4}$
$x^4+\left(1-2a\right)x^2+a^2=\frac{4a-1}{4}$
$4x^4+\left(4-8a\right)x^2+4a^2=4a-1$
$4x^4+\left(4-8a\right)x^2+4a^2-4a+1=0$
let $k = x^2$
$4k^2+\left(4-8a\right)k+4a^2-4a+1=0$
The discriminant of this system must = 0 for the circle to only touch in two places; 
$b^2-4ac=0$
$\left(4-8a\right)^2-16\left(4a^2-4a+1\right)=0$
$\left(64a^2-64a+16\right)-\left(64a^2-64a+16\right)=0$
And i think that proves the formula, but I just wouldn't know how to do that in reverse. So the question I'm asking is how would you answer;
'Give the general formula of the equation of the circle with centre $(0,a)$ which shares no more than 2 points with the curve $x^2$ '
 [/INST] 
'Give the general formula of the equation of the circle with centre $(0,a)$ which shares no more than 2 points with the curve $x^2$ '

If the equation of the circle is $x^2+(y-a)^2=r^2$, then, from $y=x^2$, we have
$$y+(y-a)^2=r^2,$$
i.e.
$$y^2+(1-2a)y+a^2-r^2=0$$
Since the discriminant has to be $0$, we have
$$(1-2a)^2-4(a^2-r^2)=0\implies r^2=\frac{4a-1}{4}$$
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