<s>[INST] Ellipse 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. 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.
 [/INST] You have done sufficient hard work.
As $m$ is the slope of the common tangent,
  we have $\displaystyle \pm \sqrt{a^2m^2+b^2} = \pm r\sqrt{1+m^2}$ 
Squaring we get $\displaystyle a^2m^2+b^2=r^2(1+m^2)$
What is $m$, compare with my other answer?
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