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(C) The tangent to the parabola $y = x^2$ at $(2, 4)$ is given by $\frac{1}{2}(y + 4) = x(2)$,which simplifies to $4x - y - 4 = 0$.Since the circle to

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

Vedclass · Answer

(C) The tangent to the parabola $y = x^2$ at $(2, 4)$ is given by $\frac{1}{2}(y + 4) = x(2)$,which simplifies to $4x - y - 4 = 0$.Since the circle touches the parabola at $(2, 4)$,the centre $(h, k)$ of the circle must lie on the normal to the parabola at $(2, 4)$.The slope of the tangent is $4$,so the slope of the normal is $-\frac{1}{4}$.The equation of the normal at $(2, 4)$ is $y - 4 = -\frac{1}{4}(x - 2)$,which simplifies to $x + 4y = 18$.Thus,$h + 4k = 18$ $(i)$.Since the circle passes through $(0, 1)$ and $(2, 4)$,the distance from the centre $(h, k)$ to these points must be equal:$(h - 2)^2 + (k - 4)^2 = (h - 0)^2 + (k - 1)^2$.Expanding this,$h^2 - 4h + 4 + k^2 - 8k + 16 = h^2 + k^2 - 2k + 1$,which simplifies to $4h + 6k = 19$ $(ii)$.Solving equations $(i)$ and $(ii)$ simultaneously: from $(i)$,$h = 18 - 4k$. Substituting into $(ii)$ gives $4(18 - 4k) + 6k = 19$,so $72 - 16k + 6k = 19$,which means $10k = 53$,so $k = \frac{53}{10}$.Then $h = 18 - 4(\frac{53}{10}) = 18 - \frac{106}{5} = \frac{90 - 106}{5} = -\frac{16}{5}$.The centre is $\left( -\frac{16}{5}, \frac{53}{10} \right)$.

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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