Consider a triangle having vertices A ( - 2, 3 ) , B ( 1, 9 ) and C...

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,

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

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

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

Circumcenter $≡ \left(\dfrac{1}{2}, \dfrac{11}{2}\right)$

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

Equation of line $L$ is

\left(y - \dfrac{11}{2}\right) = \left(\dfrac{\dfrac{11}{2} - \dfrac{17}{2}}{\dfrac{1}{2} - 2}\right)\left(x - \dfrac{1}{2}\right)$</p><p>$⇒ y = 2x + \dfrac{9}{2}$</p><p>Passing though \left(0, \dfrac{\alpha}{2}\right)$, so

$\dfrac{\alpha}{2} = \dfrac{9}{2} ⇒ \alpha = 9$