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(D) The vertex and focus of the parabola $y^{2}=2x$ are $V(0,0)$ and $S\left(\frac{1}{2}, 0\right)$ respectively.Let the equation of the circle be $(x

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(D) The vertex and focus of the parabola $y^{2}=2x$ are $V(0,0)$ and $S\left(\frac{1}{2}, 0\right)$ respectively.Let the equation of the circle be $(x-h)^{2}+(y-k)^{2}=4$.Since the circle passes through $(0,0)$,we have $h^{2}+k^{2}=4 \dots (1)$.Since the circle passes through $\left(\frac{1}{2}, 0\right)$,we have $\left(\frac{1}{2}-h\right)^{2}+k^{2}=4$,which simplifies to $h^{2}+k^{2}-h=\frac{15}{4} \dots (2)$.Subtracting $(2)$ from $(1)$,we get $h=4-\frac{15}{4}=\frac{1}{4}$.Substituting $h=\frac{1}{4}$ into $(1)$,we get $\left(\frac{1}{4}\right)^{2}+k^{2}=4$,so $k^{2}=4-\frac{1}{16}=\frac{63}{16}$,which gives $k=\pm\frac{\sqrt{63}}{4}$.For the circle to touch the parabola $y=\left(x-\frac{1}{4}\right)^{2}+\alpha$ (which has its vertex at $\left(\frac{1}{4}, \alpha\right)$),the circle must be centered at $\left(\frac{1}{4}, \frac{\sqrt{63}}{4}\right)$ and the parabola must open upwards,implying $\alpha = k + 2 = \frac{\sqrt{63}}{4} + 2$.Thus,$4\alpha = \sqrt{63} + 8$,so $4\alpha - 8 = \sqrt{63}$.Therefore,$(4\alpha-8)^{2} = 63$.

$A$ circle of radius $2$ unit passes through the vertex and the focus of the parabola $y^{2}=2x$ and touches the parabola $y=\left(x-\frac{1}{4}\right)^{2}+\alpha$,where $\alpha>0$. Then $(4\alpha-8)^{2}$ is equal to

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