The centre of the circle passing through the point (0, 1) and...

$y = {x}^{2}$

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

$⇒ 4x - y - 4 = 0$

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)$</p><p>$∴ 4 + 9 + \lambda\left(0 - 1 - 4\right) = 0$</p><p>$13 = 5\lambda ⇒ \lambda = \dfrac{13}{5}$</p><p>$∴$Circle is${x}^{2} - 4x + 4 + {y}^{2} - 8y + 16 + \dfrac{13}{5}\left(4x - y - 4\right) = 0$</p><p>$⇒ {x}^{2} + {y}^{2} + \left(\dfrac{52}{5} - 4\right)x - \left(8 + \dfrac{13}{5}\right)y + 20 - \dfrac{52}{5} = 0$</p><p>$⇒ {x}^{2} + {y}^{2} + \dfrac{32}{5}x - \dfrac{53}{5}y + \dfrac{48}{5} = 0$</p><p>$∴$Centre is \left(- \dfrac{16}{5}, \dfrac{53}{10}\right)$