In a triangle, the sum of lengths of two sides is \(x\) and the product of the lengths of the same two sides is \(y\). If \(x^2-\mathrm{c}^2-y\), where \(\mathrm{c}\) is the length of the third side of the triangle, then the circumradius of the triangle is

Let \(\mathrm{a}\) and \(\mathrm{b}\) be the lengths of two sides of \(\mathrm{a}\) triangle.
\(\therefore \quad\) According to the given condition,
\(\mathrm{a}+\mathrm{b}=x \text { and } \mathrm{ab}=y \)
\( \therefore \quad x^2-\mathrm{c}^2=y \Rightarrow(\mathrm{a}+\mathrm{b})^2-\mathrm{c}^2=\mathrm{ab} \)
\( \Rightarrow a^2+b^2+2 a b-c^2=a b \)
\( \Rightarrow a^2+b^2-c^2=-a b \)
\( \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{ab}}=-\frac{1}{2} \)
\( \Rightarrow \cos \mathrm{C}=\frac{1}{2} \)
\( \Rightarrow \mathrm{C}=\frac{2 \pi}{3} \)
\( \Rightarrow \text { circumradius }=\frac{\mathrm{c}}{2 \sin \mathrm{C}}=\frac{\mathrm{c}}{2 \sin \left(\frac{2 \pi}{3}\right)}=\frac{\mathrm{c}}{\sqrt{3}} \)