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Prove that the convex hull of a compact set $A$ in $\mathbb{C}$ is the intersection of all closed discs containing the set $A$.

One way of inclusion that convex hull is contained in the intersection of closed discs is clear. The other way, I am not able to prove. Any help please!

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    Incomplete comment: The convex hull of a compact set is a compact set. Suppose $x$ is in the intersection of closed discs containing $A$. If $x \in Conv(A)$ then we are done. Else, $x$ is not in the compact and convex set $Conv(A)$ and so we can invoke a strict separation theorem. Intuitively, we can then make a very large disc (with a huge radius) that contains $Conv(A)$ and is in the lower half of the separation (to get a contradiction). – 2017-02-20
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    @Michael What separation theorem can we use? Can you make it more precise? – 2017-02-20
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    The hyperplane separation theorem. In $\mathbb{R}^2$ it says that if we have a point $x$ not in the compact convex set $Conv(A)$, then there exists a line that neither intersects $x$ nor $Conv(A)$, and contains $x$ on one side, $Conv(A)$ on the other. – 2017-02-20
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    @Michael Thanks – 2017-02-20

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