Let C be a circle passing through the points A(2, $-$1) and B(3, 4). The line segment AB s not a diameter of C. If r is the radius of C and its centre lies on the circle ${(x - 5)^2} + {(y - 1)^2} = {{13} \over 2}$, then r² is equal to :

<p>Equation of perpendicular bisector of AB is</p> <p>$$y - {3 \over 2} = - {1 \over 5}\left( {x - {5 \over 2}} \right) \Rightarrow x + 5y = 10$$</p> <p>Solving it with equation of given circle,</p> <p>${(x - 5)^2}{\left( {{{10 - x} \over 5} - 1} \right)^2} = {{13} \over 2}$</p> <p>$\Rightarrow {(x - 5)^2}\left( {1 + {1 \over {25}}} \right) = {{13} \over 2}$</p> <p>$\Rightarrow x - 5 = \pm \,{5 \over 2} \Rightarrow x = {5 \over 2}$ or ${{15} \over 2}$</p> <p>But $x \ne {5 \over 2}$ because AB is not the diameter.</p> <p>So, centre will be $\left( {{{15} \over 2},{1 \over 2}} \right)$</p> <p>Now, $${r^2} = {\left( {{{15} \over 2} - 2} \right)^2} + {\left( {{1 \over 2} + 1} \right)^2}$$</p> <p>$= {{65} \over 2}$</p>