The line x=2y intersects the ellipse frac{x^{ | vedclass

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(D) Given equations are $x=2y$ $(i)$ and $\frac{x^{2}}{4}+y^{2}=1$ (ii).Substituting $x=2y$ into (ii),we get:$\frac{(2y)^{2}}{4}+y^{2}=1$$\frac{4y^{2}

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$x^{2}+y^{2}=\frac{5}{2}$

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(D) Given equations are $x=2y$ $(i)$ and $\frac{x^{2}}{4}+y^{2}=1$ (ii).Substituting $x=2y$ into (ii),we get:$\frac{(2y)^{2}}{4}+y^{2}=1$$\frac{4y^{2}}{4}+y^{2}=1$$y^{2}+y^{2}=1$$2y^{2}=1$ $\Rightarrow y^{2}=\frac{1}{2}$ $\Rightarrow y=\pm \frac{1}{\sqrt{2}}$.From $(i)$,$x=2y$,so $x=\pm 2(\frac{1}{\sqrt{2}}) = \pm \sqrt{2}$.Thus,the points of intersection are $P(\sqrt{2}, \frac{1}{\sqrt{2}})$ and $Q(-\sqrt{2}, -\frac{1}{\sqrt{2}})$.The equation of a circle with diameter $PQ$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.$(x-\sqrt{2})(x+\sqrt{2}) + (y-\frac{1}{\sqrt{2}})(y+\frac{1}{\sqrt{2}}) = 0$$(x^{2}-2) + (y^{2}-\frac{1}{2}) = 0$$x^{2}+y^{2} = 2 + \frac{1}{2}$$x^{2}+y^{2} = \frac{5}{2}$.

The line $x=2y$ intersects the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at the points $P$ and $Q$. The equation of the circle with $PQ$ as diameter is

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