Problem :
Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r.
Few concepts about Ellipse :
Equation of Tangent to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point $(x_1,y_1)$ is given by $T = \frac{xx_1}{a^2} +\frac{yy_1}{b^2}-1$
Also point of contact where line $y =mx +c $ touched the ellipse .
The line is y =mx +c touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ when $c =\pm \sqrt{a^2m^2+b^2}$
$(x_1,y_1) =(\pm \frac{a^2m}{\sqrt{a^2m^2 +b^2}}, \pm \frac{b^2}{\sqrt{a^2m^2+b^2}})$
Now, we know that equation of any tangent to the given ellipse is $y =mx \pm \sqrt{a^2m^2+b^2}$
Now if it touches $x^2+y^2=r^2$
Then
$ \sqrt{a^2m^2+b^2} = r\sqrt{1+m^2}$ I am unable to understand this condition ..please guide for this particular condition only… thanks.
Best Answer
If we use parametric forms of the ellipse $(a\cos\phi,b\sin\phi)$ and of the circle $(r\cos\psi,r\sin\psi)$
we get the tangents to be $\displaystyle x\frac{\cos\phi}a+y\frac{\sin\phi}b=1\iff x b\cos\phi+y a\sin\phi=ab$
and for the circle $x\cos\psi+y\sin\psi=r$ with slope $\displaystyle-\frac{\cos\psi}{\sin\psi}=-\cot\psi\ \ \ \ (1)$
These two equations have to be identical implies $$\frac{b\cos\phi}{\cos\psi}=\frac{a\sin\phi}{\sin\psi}=\frac{ab}r$$
So, $\displaystyle\cos\phi=\frac{a\cos\psi}r,\sin\phi=\frac{b\sin\psi}r$
Squaring & adding we get, $\displaystyle(a\cos\psi)^2+(b\sin\psi)^2=r^2$
Divide either sides by $\sin^2\psi,$
$\displaystyle(a\cot\psi)^2+b^2=r^2(1+\cot^2\psi)\implies \cot^2\psi=\frac{r^2-b^2}{a^2-r^2}$ which must be $\ge0$
$\displaystyle\implies(r^2-b^2)(a^2-r^2)\ge0\iff (r^2-a^2)(r^2-b^2)\le0$
$\displaystyle\implies $ min$(a,b)\le r\le$max$(a,b)$
The value of $r$ will dictate the value of the slope in $(1)$