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(D) The equation of a tangent to the circle $x^2+y^2=16$ with slope $m$ is $y=mx \pm 4\sqrt{1+m^2}$.The equation of a tangent to the ellipse $\frac{x^

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$\sqrt{11}y=2x+4\sqrt{15}$

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(D) The equation of a tangent to the circle $x^2+y^2=16$ with slope $m$ is $y=mx \pm 4\sqrt{1+m^2}$.The equation of a tangent to the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ with slope $m$ is $y=mx \pm \sqrt{49m^2+4}$.For the tangents to be common,the constant terms must be equal:$16(1+m^2) = 49m^2+4$$16+16m^2 = 49m^2+4$$33m^2 = 12$$m^2 = \frac{12}{33} = \frac{4}{11}$$m = \pm \frac{2}{\sqrt{11}}$.Substituting $m^2 = \frac{4}{11}$ into the circle's tangent equation:$y = \pm \frac{2}{\sqrt{11}}x \pm 4\sqrt{1+\frac{4}{11}} = \pm \frac{2}{\sqrt{11}}x \pm 4\sqrt{\frac{15}{11}} = \pm \frac{2}{\sqrt{11}}x \pm \frac{4\sqrt{15}}{\sqrt{11}}$.Multiplying by $\sqrt{11}$,we get $\sqrt{11}y = \pm 2x \pm 4\sqrt{15}$.Comparing with the options,the correct equation is $\sqrt{11}y=2x+4\sqrt{15}$.

The equation of a common tangent to the circle $x^2+y^2=16$ and the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ is

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