The radius of circle, touching the parabola y^ 2 = 8xat (2,4) and...

Equation of the tangent at $\left(2,4\right)$ on the parabola $y^{2} = 8x$ is $y\left(4\right) = 8\left(\dfrac{x + 2}{2}\right) ⇒ y = x + 2$
Let equation of circle touching line $y = x + 2$ at \left(2,4\right)$is${\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} + \lambda\left(x - y + 2\right) = 0$which passes through \left(0,4\right)⇒ 4 + 0 + \lambda\left(0 - 4 + 2\right) ⇒ \lambda = 2$
$⇒$Equation of the required circle is $x^{2} + y^{2} - 2x - 10y + 24 = 0$
$⇒$~The radius of the circle is $\sqrt{{1}^{2} + {5}^{2} - 24} = \sqrt{2}$