The number of integral points on the circle, touching the parabola $y^2=8x$ at $\left(2,4\right)$ and passing through $\left(0,4\right),$ are equal to

Equation of tangent at $\left(\mathrm{2,4}\right)$ on the parabola $y^2=8x$ is

$y\left(4\right)=8\left(\frac{x+2}{2}\right)\Rightarrow y=x+2$
Let the equation of the circle touching line $y=x+2$ at $\left(\mathrm{2,4}\right)$ is

${\left(x-2\right)}^2+{\left(y-4\right)}^2+\lambda \left(x-y+2\right)=0$ which passes through $\left(0,4\right)$

$\Rightarrow 4+0+\lambda \left(0-4+2\right)\Rightarrow \lambda =2$
$\Rightarrow$ Required circle is $x^2+y^2-2x-10y+24=0$
$\Rightarrow {\left(x-1\right)}^2+{\left(y-5\right)}^2=2$
If $x$ and $y$ are integers, then

${\left(x-1\right)}^2=1={\left(y-5\right)}^2$
$\Rightarrow x=0,2$ and $y=4,6$
$\Rightarrow 4$ integral points lie on the circle