In a triangle,two vertices are (2, 3) and (4, | vedclass

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(D) Let the vertices be $A(2, 3)$ and $B(4, 0)$. Let the circumcentre be $O(2, z)$.Since $O$ is the circumcentre,the distance from $O$ to all vertices

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$\frac{13}{6}$

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(D) Let the vertices be $A(2, 3)$ and $B(4, 0)$. Let the circumcentre be $O(2, z)$.Since $O$ is the circumcentre,the distance from $O$ to all vertices is equal to the circumradius $R$.Thus,$OA^2 = OB^2$.$OA^2 = (2-2)^2 + (z-3)^2 = (z-3)^2$$OB^2 = (4-2)^2 + (0-z)^2 = 2^2 + z^2 = 4 + z^2$Equating $OA^2 = OB^2$:$(z-3)^2 = 4 + z^2$$z^2 - 6z + 9 = 4 + z^2$$-6z = 4 - 9$$-6z = -5$$z = \frac{5}{6}$Now,calculate the circumradius $R = OA = \sqrt{(2-2)^2 + (z-3)^2} = |z-3|$.$R = |\frac{5}{6} - 3| = |\frac{5-18}{6}| = |-\frac{13}{6}| = \frac{13}{6}$.

In a triangle,two vertices are $(2, 3)$ and $(4, 0)$,and its circumcentre is $(2, z)$ for some real number $z$. The circumradius is

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