I have two equations:
(1) $y=0$ and (2) $x^2 + y^2 = 2^2$
I also have another equation: (3) $x^4 + (y-a)^4 = b^4$
$a$ and $b$ are chosen such that (3) intersects with (1) at $x=y=0$ and intersects with (2) at exactly two points and at these points (2) and (3) are tangential . I do not know how to approach finding these two points. Can someone please give me a clue how to get started?
If (3) intersects (1) at $x=y=0$, that says $a^4 = b^4$. Since this is "geometry", I'm assuming $a$ and $b$ are real, so either $a=b$ or $a=-b$. Since (3) is unchanged by replacing $b$ by $-b$, I'll assume $a=b$.
The resultant of $x^2 + y^2 - 2^2$ and $x^4 + (y-b)^4 - b^4$ with respect to $y$ is $$(16 -4 b^3 y + (6 b^2-8) y^2 - 4 b y^3 + 2 y^4 )^2$$ Thus an intersection of (2) and (3) has $y$ satisfying $g(y) = 16 -4 b^3 y + (6 b^2-8) y^2 - 4 b y^3 + 2 y^4 = 0$. Each $y$ satisfying this will correspond to two points $(x,y)$ and $(-x,y)$. In order for the two curves to be tangential, the discriminant of $g(y)$ must be $0$. That will give you a polynomial in $b$ of degree $12$ (but even, so of degree $6$ in $b^2$), which has two real roots: approximately $b = \pm 0.9720586597$. This sextic has Galois group $S_6$, so no solutions in radicals.