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(B) The equation of the parabola is $y=x^{2}$.The slope of the tangent at point $P(2,4)$ is given by $\left.\frac{dy}{dx}\right|_{(2,4)} = 2x|_{x=2} =

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

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(B) The equation of the parabola is $y=x^{2}$.The slope of the tangent at point $P(2,4)$ is given by $\left.\frac{dy}{dx}\right|_{(2,4)} = 2x|_{x=2} = 4$.The equation of the tangent line at $(2,4)$ is $(y-4) = 4(x-2)$,which simplifies to $4x-y-4=0$.The family of circles touching the parabola at $(2,4)$ is given by $(x-2)^{2} + (y-4)^{2} + \lambda(4x-y-4) = 0$.Since the circle passes through $(0,1)$,we substitute $x=0$ and $y=1$ into the equation:$(0-2)^{2} + (1-4)^{2} + \lambda(4(0)-1-4) = 0$$4 + 9 - 5\lambda = 0$ $\Rightarrow 13 = 5\lambda$ $\Rightarrow \lambda = \frac{13}{5}$.Substituting $\lambda = \frac{13}{5}$ into the circle equation:$(x-2)^{2} + (y-4)^{2} + \frac{13}{5}(4x-y-4) = 0$$x^{2} - 4x + 4 + y^{2} - 8y + 16 + \frac{52}{5}x - \frac{13}{5}y - \frac{52}{5} = 0$$x^{2} + y^{2} + (\frac{52}{5} - 4)x - (8 + \frac{13}{5})y + (20 - \frac{52}{5}) = 0$$x^{2} + y^{2} + \frac{32}{5}x - \frac{53}{5}y + \frac{48}{5} = 0$.The centre of the circle $x^{2} + y^{2} + 2gx + 2fy + c = 0$ is $(-g, -f)$.Here,$2g = \frac{32}{5} \Rightarrow g = \frac{16}{5}$ and $2f = -\frac{53}{5} \Rightarrow f = -\frac{53}{10}$.Thus,the centre is $(-\frac{16}{5}, \frac{53}{10})$.

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

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