Consider a triangle having vertices A ( - 2 , 3 ) ,   B ( 1 , 9 ) and C ( 3 ,   8 ) . If a line L…

We have,

$AB=\sqrt{{\left(-2-1\right)}^2+{\left(3-9\right)}^2}=\sqrt{45}$

$AC=\sqrt{{\left(-2-3\right)}^2+{\left(3-8\right)}^2}=\sqrt{50}$

$BC=\sqrt{{\left(3-1\right)}^2+{\left(8-9\right)}^2}=\sqrt{5}$

Here,

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow (\sqrt{50}{)}^2=(\sqrt{45}{)}^2+(\sqrt{5}{)}^2$

Hence, $\angle B=90°$, so circumcentre is the mid-point of hypotenuse i.e., $AC$.

Circumcenter $\equiv \left(\frac{1}{2},\frac{11}{2}\right)$

Mid-point of $BC=\left(2,\frac{17}{2}\right)$

Equation of line $L$ is

$\left(y-\frac{11}{2}\right)=\left(\frac{{\displaystyle \frac{11}{2}}-{\displaystyle \frac{17}{2}}}{{\displaystyle \frac{1}{2}}-2}\right)\left(x-\frac{1}{2}\right)$

$\Rightarrow y=2x+\frac{9}{2}$

Passing though $\left(0,\frac{\alpha }{2}\right)$, so

$\frac{\alpha }{2}=\frac{9}{2}\Rightarrow \alpha =9$