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(D) The equation of the tangent to the curve $y=x^2$ at $(2,4)$ is given by $\frac{y+4}{2} = 2x$,which simplifies to $4x - y - 4 = 0$. The equation of

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

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(D) The equation of the tangent to the curve $y=x^2$ at $(2,4)$ is given by $\frac{y+4}{2} = 2x$,which simplifies to $4x - y - 4 = 0$. The equation of a circle touching this curve at $(2,4)$ with the tangent $4x - y - 4 = 0$ is $(x-2)^2 + (y-4)^2 + \lambda(4x - y - 4) = 0$. Since the circle passes through $(0,1)$,we substitute these coordinates: $(0-2)^2 + (1-4)^2 + \lambda(0-1-4) = 0$. This gives $4 + 9 - 5\lambda = 0$,so $\lambda = \frac{13}{5}$. Substituting $\lambda$ back into the circle equation: $x^2 - 4x + 4 + y^2 - 8y + 16 + \frac{52}{5}x - \frac{13}{5}y - \frac{52}{5} = 0$. Simplifying,we get $x^2 + y^2 + x(\frac{52}{5} - 4) - y(8 + \frac{13}{5}) + (20 - \frac{52}{5}) = 0$. The centre of the circle is $(-\frac{1}{2}(\frac{52}{5} - 4), \frac{1}{2}(8 + \frac{13}{5})) = (-\frac{16}{5}, \frac{53}{10})$.

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

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