A circle passes through the points 2 , 3 and 4 , 5 . If its centre lies on the line y - 4 x + 3 = 0 , then…

Let centre of circle be $\mathrm{C}\left(\alpha , \beta \right)$.

It lies in line $y-4x+3-0\Rightarrow \beta =4\alpha -3$

$\therefore C \left(\alpha , 4\alpha -3\right)$

$\therefore CA=CB$

By distance formula,

$\Rightarrow {\left(\alpha -2\right)}^2+{\left(4\alpha -6\right)}^2={\left(\alpha -4\right)}^2+{\left(4\alpha -8\right)}^2$

$\Rightarrow -4\alpha +4-48\alpha +36=-8\alpha +16-64\alpha +64$

$\Rightarrow \left(64+8-4-48\right)\alpha =80-40$

$\Rightarrow \alpha =\frac{40}{20}=2$

$\therefore C\left(2,5\right)$

$\therefore r=\sqrt{{\left(2-2\right)}^2+{\left(5-3\right)}^2}=2$