Let A α , - 2 , B α , 6 and C α 4 , - 2 be vertices of a ∆ A B C . If 5 , α 4 is…

Given,

$A\left(\mathrm{\alpha },-2\right), B\left(\mathrm{\alpha },6\right) \& C\left(\frac{\alpha }{4},-2\right)$

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

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

So, $∆ABC$ is right angled at $A$.

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

Now given circumcentre $\equiv \left(5,\frac{\alpha }{4}\right)$

So on comparing we get,$\frac{5\alpha }{8}=5 \& \frac{\alpha }{4}=2$

$\Rightarrow \alpha =8$

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

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

Perimeter $=24$

Circumradius $=\frac{10}{2}=5$

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