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Q.
Equation of a common tangent to the circle $x^{2}+y^{2}=4$ and to the ellipse $2 x^{2}+25 y^{2}=50$ is
TS EAMCET 2020
Solution:
Let $y=m x+c$ is common tangent of circle
$x^{2}+y^{2}=4$ and ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{2}=1$
$\therefore y=m x \pm 2 \sqrt{1+m^{2}}$
and $y=m x \pm \sqrt{25 m^{2}+2}$ are coincide
$\because 4\left(1+m^{2}\right)=25 m^{2}+2$
$21 m^{2}=2 $
$\Rightarrow m=\pm \sqrt{\frac{2}{21}}$
therefore Equation of tangent is
$y=\frac{\sqrt{2}}{\sqrt{21}} x \pm \frac{2 \sqrt{23}}{\sqrt{21}} $
$\Rightarrow \sqrt{2 x}-\sqrt{21} y+2 \sqrt{23}=0$