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Suppose I have a circumcircle $C$ about the three points $x_1, x_2, x_3$ in the plane. Now I have a fourth point $x_4$ that lies in the interior of $C$, and is on the side of the segment $x_2 x_3$ opposite from $x_1$. Does it necessarily follow that the circumcircle about $x_1, x_2, x_4$ must have a smaller radius than $C$?

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If $\widehat{x_2 x_1 x_3}\leq\frac{\pi}{2}$ you have a relation between the two circumradii, and the result follows from the law of sines. In any triangle you have: $$ 2R = \frac{a}{\sin \widehat{A}}, $$ so, if you call $a=\overline{x_2 x_3},\; \widehat{A}=\widehat{x_2 x_1 x_3},\; \widehat{C}=\widehat{x_3 x_4 x_2}\;$, you have: $$ \pi-\widehat{A}<\widehat{C}\leq\pi, $$ so: $$ \sin\widehat{C}<\sin\widehat{A}, $$ then the circumradius of $x_2 x_3 x_4$ is bigger than the circumradius of $x_1 x_2 x_3$.

If $\widehat{x_2 x_1 x_3}\leq\frac{\pi}{2}$ does not hold, you cannot say anything: consider the case in which $x_1,x_4$ are symmetric with respect to the line $x_2 x_3$.

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