Tangent to $x^2+y^2-6x-6y=-13$ and $x^2+y^2+2x+2y=-1$ Considering the circles $\lambda: x^2+y^2-6x-6y=-13$ and $\theta: x^2+y^2+2x+2y=-1$ find the line simultaneously tangent to them.
I found the implicit derivative of those two,
$\lambda: y'=-\frac{x-3}{y-3}$ and $\theta: y'=-\frac{x+1}{y+1}$ but then I don't know how to proceed.
I thought that setting $y'_{\lambda}=y'_{\theta}$ would solve, but it didn't.
If anyone could help I'll appreciate.
 A: 
To prove that the point of intersection of the two direct common tangents lies on the line joining the centers of the two circles, notice that angles $\angle ACD$, $\angle AFE$, $\angle BDG$ and $\angle BEG$ are all right angles. Thus, $\angle CAB$+ $\angle DBA$ = $\angle FAB$ + $\angle ABE$ = $180^{\circ}$.  Also, $\angle GBD$ + $\angle DBG$ = $\angle GBD$ + $\angle CAB$ = $90^{\circ}$. Now conclude that $\angle DBG$ + $\angle DBA$ = $180^{\circ}$. Done.  Now, notice that by similarity of $\Delta CAG$ and $\Delta DBG$, $$\frac{AG}{BG}=\frac{AC}{BD}=\frac{R}{r}$$
 By the external section formula, we have $$G\equiv \frac{R(-1,-1)-r(3,3)}{R-r}=(\frac{-R-3r}{R-r}, \frac{-R-3r}{R-r})$$ Now, the line may be written as $\displaystyle y+\frac{R+3r}{R-r}=m(x +\frac{R+3r}{R-r})$.
 The distance of a point $(h,k)$ from a line $ax+by+c=0$ is $\left\vert \dfrac{ah+bk+c}{a^2+b^2}\right\vert$.
The distance from any center to the tangent is equal to the radius of that circle so write:$$\left\vert\dfrac{-1-m(-1)+(1-m) \frac{R+3r}{R-r}}{\sqrt{1+m^2}}\right\vert= 1$$ Now you will get two values of m and consequently, two common tangents. Oh, I forgot to add: R = $\sqrt 5$ and r=1.

Now I hope you will be able to find the transverse common tangents by yourself.
A: You are waving the fact that the two circumferences are different loci:
the tangent point on one will not have the same $(x,y)$ as on the other.
If you want to proceed on that way you shall write
$$
\begin{array}{l}
 \left\{ \begin{array}{l}
 z_1  =  - 13 = f_1 \left( {x_1 ,y_1 } \right) \\ 
 z_1  =  - 1 = f_2 \left( {x_2 ,y_2 } \right) \\ 
 dz_1  = 0 = \frac{\partial }{{\partial x_1 }}f_1 \left( {x_1 ,y_1 } \right)dx_1  + \frac{\partial }{{\partial y_1 }}f_1 \left( {x_1 ,y_1 } \right)dy_1  \\ 
 dz_2  = 0 = \frac{\partial }{{\partial x_2 }}f_2 \left( {x_2 ,y_2 } \right)dx_2  + \frac{\partial }{{\partial y_2 }}f_2 \left( {x_2 ,y_2 } \right)dy_2  \\ 
 \end{array} \right.\quad  \Rightarrow  \\ 
 \quad  \Rightarrow \left\{ \begin{array}{l}
 z_1  =  - 13 = f_1 \left( {x_1 ,y_1 } \right) \\ 
 z_1  =  - 1 = f_2 \left( {x_2 ,y_2 } \right) \\ 
 \frac{{dy_1 }}{{dx_1 }} =  - \frac{{\frac{\partial }{{\partial x_1 }}f_1 \left( {x_1 ,y_1 } \right)dx_1 }}
{{\frac{\partial }{{\partial y_1 }}f_1 \left( {x_1 ,y_1 } \right)}} \\ 
 \frac{{dy_2 }}{{dx_2 }} =  \cdots  \\ 
 \end{array} \right. \\ 
 \end{array}
$$
A: The dual conic method; a method that works for finding the common tangents of any two smooth conics.
$x^2+y^2-6x-6y=-13$ has dual $4X^2+18XY+6Y+4X^2+6X+1=0$
$x^2+y^2+2x+2y=-1$ has dual $2XY-2Y-2X+1=0$
The intersection points in the dual plane from i.e. the grobner basis $\langle 4Y^4+16Y^3-24Y^2+5, X+2Y^3+10Y^2-2Y-3\rangle$ are $P_1: (X,Y)=(-0.4090941234707588,0.6451620997691458),\\P_2: (X,Y)=(0.6451620997691458,-0.4090941234707588),\\P_3: (X,Y)=(0.9187629817678283,-5.154831199068685),\\P_4: (X,Y)=(-5.154831199068685,0.9187629817678283)$
corresponding to the common tangents $X(P_i)x+Y(P_i)y+1=0, i=1\ldots 4$ in the usual plane.
