Common tangent to a circle and ellipse Hey guys i am noy able to solve this problem.So please do help me in solving this.The equation of common tangent to ellipse 
\begin{equation*}
x^2 +2y^2=1
\end{equation*}
and circle 
\begin{equation*}
x^2 +y^2=\frac{2}{3}
\end{equation*}
is?

The equation of tangent on the ellipse is given by:
$$x x_1 + 2y y_1 = 1$$
The perpendicular distance of the line from the centre of circle (0, 0)
$= \frac 1{\sqrt{x_1^2 + 4y_1^2}}$
This distance is equal to the radius of the circle as the line should also be tangent to the circle:
$$\frac 1{\sqrt{x_1^2 + 4y_1^2}} = \sqrt \frac23$$
Since $(x_1, y_1)$ lie on the ellipse, $x_1^2 + 2y_1^2 = 1$; These give $y_1 = \pm \frac12$ and $x_1 = \pm \sqrt \frac12$.
Therefore, the tangents are: $$ \frac{\pm x}{\sqrt 2} \pm y = 1$$

Way 2: Let the common tangent be $y = mx + c$.
It meets the ellipse at only one point:
$$mx + c = \sqrt{\frac{1-x^2}2}$$
$$(2m^2 + 1)x^2 + 4mcx + 2c^2 - 1 = 0$$
This quadratic has equal roots so discriminant = 0
$$2m^2 + 1 = 2c^2$$
Similarly for the circle,
$$mx + c = \sqrt{\frac23 - x^2}$$
Simplifying and setting discriminant to 0 gives:
$$2m^2 + 2 = 3c^2$$
Solving for m and c gives
$$ \frac{\pm x}{\sqrt 2} \pm y = 1$$
Desmos graph
