Let A (α, - 2), B (α, 6) and C((α)/(4), - 2) be vertices of a ∆ ABC...

Given,

$A\left(\alpha, - 2\right), B\left(\alpha, 6\right) \text{ and } C\left(\dfrac{\alpha}{4}, - 2\right)$

Now slope of $AC = 0$ and slope of $AB = \dfrac{8}{0} \to \infty$

Now we can see $AC$ is perpendicular to $AB$.

So, $∆ ABC$ is right angled at $A$.

So, Circumcentre =mid point of $BC = \left(\dfrac{5\alpha}{8}, 2\right)$

Now given circumcentre $≡ \left(5, \dfrac{\alpha}{4}\right)$

So on comparing we get,$\dfrac{5\alpha}{8} = 5 \text{ and } \dfrac{\alpha}{4} = 2$

$⇒ \alpha = 8$

Now plotting the diagram and finding the sides with distance formula we get,

Area $= \dfrac{1}{2}\left(6\right)\left(8\right) = 24$

Perimeter $= 24$

Circumradius $= \dfrac{10}{2} = 5$

Inradius $= \dfrac{\Delta}{s} = \dfrac{24}{12} = 2$