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(B) Let the equation of the common tangent be $y=mx+c$. For the circle $x^2+y^2=4$,the condition for tangency is $c^2 = r^2(1+m^2)$,where $r^2=4$. So,

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$\sqrt{2}x-\sqrt{21}y+2\sqrt{23}=0$

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(B) Let the equation of the common tangent be $y=mx+c$. For the circle $x^2+y^2=4$,the condition for tangency is $c^2 = r^2(1+m^2)$,where $r^2=4$. So,$c^2 = 4(1+m^2)$. For the ellipse $\frac{x^2}{25} + \frac{y^2}{2} = 1$,the condition for tangency is $c^2 = a^2m^2 + b^2$,where $a^2=25$ and $b^2=2$. So,$c^2 = 25m^2 + 2$. Equating the two expressions for $c^2$: $4(1+m^2) = 25m^2 + 2$ $4 + 4m^2 = 25m^2 + 2$ $21m^2 = 2$ $\Rightarrow m^2 = \frac{2}{21}$ $\Rightarrow m = \pm \sqrt{\frac{2}{21}}$. Substituting $m^2$ back into $c^2 = 4(1+m^2)$: $c^2 = 4(1 + \frac{2}{21}) = 4(\frac{23}{21}) = \frac{92}{21}$. Thus,$c = \pm \sqrt{\frac{92}{21}} = \pm \frac{2\sqrt{23}}{\sqrt{21}}$. The equation of the tangent is $y = \pm \sqrt{\frac{2}{21}}x \pm \frac{2\sqrt{23}}{\sqrt{21}}$. Multiplying by $\sqrt{21}$: $\sqrt{21}y = \pm \sqrt{2}x \pm 2\sqrt{23}$. Rearranging gives $\sqrt{2}x - \sqrt{21}y + 2\sqrt{23} = 0$.

The equation of a common tangent to the circle $x^2+y^2=4$ and the ellipse $2x^2+25y^2=50$ is

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