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The problem I'm trying to solve is : Given a circle of equation $x^2+y^2=4$ ,an ellipse of equation $2x^2+5y^2=10$ and their mutual tangent whose equation is $y=kx+n$, determine $k^2+n^2$. I would like some kind of a subtle hint, not a complete solution. My attempt was to use equations of tangents line for circle and ellipse but the system of equations I get that way doesn't really help me in any way.

Thanks ;)

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Describe with an equation the family of the lines that are tangent to the circle. Hint for this: the line is perpendicular to the radius, being $\alpha$ the angle between the radius and $X$ axis. Don't be afraid to use trigonometry.

Make a system of equations: an equation is that of the line (it depends on $\alpha$), and the other are that of ellipse. The discriminant of the system must be $0$. Find $\alpha$, or its cosine.

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Since the tangent line has to pass through the circle as well as ellipse it should intersect both of them at coinciding points (equal roots). Thus if you look at circle line intersection then you can get $$x^2+(kx+n)^2=4.$$ This is a quadratic in $x$ and it should have two equal roots (for tangency). Thus the discriminant should be $0$. Now do the same for ellipse.

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Thinking analytically,

the equation of tangent at $(2\cos\alpha,2\sin\alpha)$ of the circle is $$\displaystyle x(2\cos\alpha)+y(2\sin\alpha)=4\iff x(\cos\alpha)+y(\sin\alpha)=2\ \ \ \ (1)$$

The equation of tangent at $(\sqrt5\cos\beta,\sqrt2\sin\beta)$ of the ellipse is $$\displaystyle \frac{x(\sqrt5\cos\beta)}5+\frac{y(\sqrt2\sin\beta)}2=1\iff x(2\sqrt5\cos\beta)+y(5\sqrt2\sin\beta)=10\ \ \ \ (2)$$

For common tangent, $(1),(2)$ should be same, i.e., $$\frac{\cos\alpha}{2\sqrt5\cos\beta}=\frac{\sin\alpha}{5\sqrt2\sin\beta}=\frac2{10}$$


Squaring & Adding we get $$25=20\cos^2\beta+50\sin^2\beta\iff5=4\cos^2\beta+10\sin^2\beta=4(\cos^2\beta+\sin^2\beta)+6\sin^2\beta$$ $$\iff6\sin^2\beta=5-4=1\ \ \ \ (3)$$

Express $(2)$ as $\displaystyle y=kx+n$ and find $k,n$ in terms of $\beta$ & use $(3)$

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