Let $S=A\cup B$ where $A=\{(x_1,x_2):x_1<0,x_1^2+x_2^2\le4\}$ and $B=\{(x_1,x_2):x_1\ge0,-2\le x_2 \le2\}. $ Find the support function.
The support function is defined as $f(y)=\text{sup} \{y^tx:x\in S\},$ where $S$ is a convex and bounded set in $\mathbb R^n$ and $f:\mathbb R^n\to\mathbb R.$
From here, $$f(y)=\text{sup} \{y^t(x_1,x_2):(x_1,x_2)\in A\cup B\}=\text{sup} \{(y_1x_1+x_2y_2):(x_1,x_2)\in A\cup B\}$$
What do I do from here?
Should I consider cases for $(x_1,x_2)?$ i.e. if $(x_1,x_2)\in A$ or in B or in both?
The dot product $y^t x$ equals $\cos \theta |y||x|$ where $\theta$ is the smaller angle between the two vectors. So to maximize $y^t x$ you choose a vector $x$ that is parallel to $y$ since then $\cos (0) = 1 = $ the maximum value $\cos$ can take.
Thus the $x$ maximizing the orange $y$ vector pictured is always going to be on the boundary of $S$. Thus $f(y)$ is simply the magnitude $|y| \equiv \sqrt{y^t y}$ times the distance from the origin to where the boundary of $S$ intersects the ray along $y$.
It seems weird to me that you'd have to compute the actual expression for that, which will usually be some conditionals for the square on the right half plane, and a little simpler for the circle on the left half plane.
If you have to compute the explict description of it, you can find where the ray $r(t) = y t$ intersects each line in the square and similarly where it intersects the circle.