Let C be a circle passing through the points A(2, - 1) and B(3, 4)...

Equation of line joining $AB$ is $</p><p>$y + 1 = 5 \left(x - 2\right)$</p><p>i.e.$5x - y - 11 = 0$</p><p>Also$AB = \sqrt{{\left(3 - 2\right)}^{2} + {\left(4 + 1\right)}^{2}} = \sqrt{26}$</p><p>$r$is the radius of the circle$C$</p><p>For the circle${\left(x - 5\right)}^{2} + {\left(y - 1\right)}^{2} = \dfrac{13}{2}

centre is \left(5, 1\right)$and radiusR = \sqrt{\dfrac{13}{2}}$

Equation of perpendicular bisector of $AB$ will be $y - 1 = -$\dfrac{1}{5}\left(x - 5\right) ⇒ x + 5y - 10 = 0$</p><p>Solving this equation of line with${\left(x - 5\right)}^{2} + {\left(y - 1\right)}^{2} = \dfrac{13}{2}, we get

$M$ as \left(\dfrac{5}{2}, \dfrac{3}{2}\right)$</p><p>So$M$lies on the line joining$AB$.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/baa60d44-f066-40d7-a9db-3b7768ecc3c6-image.png" style="width: 300px; height: 303px;" />$

Now ${r}^{2} = C{M}^{2} + A{M}^{2}$

$= {\left(2 \times \sqrt{\dfrac{13}{2}}\right)}$^{2} + {\left(\sqrt{\dfrac{13}{2}}\right)}^{2}

${r}^{2} =$\dfrac{65}{2}