A circle passes through the points (2, 3) and (4, 5) . If its...

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

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

$∴ C\left(\alpha, 4\alpha - 3\right)$

$∴ CA = CB$

By distance formula,

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

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

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

$⇒ \alpha = \dfrac{40}{20} = 2$

$∴ C\left(2, 5\right)$

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