The centre of the circle passing through the point $\left(0, 1\right)$ and touching the parabola $y=x^2$ at the point $\left(2, 4\right)$ is :

$y=x^2$

Tangent at $\left(2,4\right)$ is $\frac{1}{2}\left(y+4\right)=2x$, using Cartesian form.

$\Rightarrow 4x-y-4=0$

Equation of circle

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

It passes through $\left(0,1\right)$

$\therefore 4+9+\lambda \left(0-1-4\right)=0$

$13=5\lambda \Rightarrow \lambda =\frac{13}{5}$

$\therefore$Circle is $x^2-4x+4+y^2-8y+16+\frac{13}{5}\left(4x-y-4\right)=0$

$\Rightarrow x^2+y^2+\left(\frac{52}{5}-4\right)x-\left(8+\frac{13}{5}\right)y+20-\frac{52}{5}=0$

$\Rightarrow x^2+y^2+\frac{32}{5}x-\frac{53}{5}y+\frac{48}{5}=0$

$\therefore$ Centre is $\left(-\frac{16}{5},\frac{53}{10}\right)$