This is similar to the question posed here, but I am interested in the case of two circles $C_1$ and $C_2$ of different radii $r_1$ and $r_2$, sharing a common tangent. The circles are not co-planar. I know the centers $c_1, c_2$ of the circles.
I am not sure how many possible tangent planes common to both can be constructed where the planes I am looking for do not contain the common tangent at the point of contact. The question linked says that for three circles in $\mathbb{R}^3$, it is atmost 8, assuming no common tangents and such degeneracies. This is probably the same as asking what are the possible resting planes tangential to both the circles and which contain both the circles on the same side. The planes through the common tangent do not satisfy this requirement.
My intuition or rather guess is that in the present case with two circles tangential at a point, it should not be more than 4 including the plane containing the common point. But is there a way to prove this rigorously?
Lastly, how do I find the equation of the planes?
I was thinking of constructing two cones that fit the two circles and whose axes pass through the respective centres of the circles and are thus normal to the planes in which the circles lie. These normals intersect at the common vertex of the two cones. Is this construction even unique? I am not sure if this is right approach or how to build on this if so?
The text of the question is not very clear, but I gather from the comments of the OP that, given two tangent circles in 3D, we want to construct two cones with the same vertex having those circle as base, and then the planes through the vertex tangent to both cones.
In the figure below I represented the given circles (with centers $O$, $O'$ and radii $r$ $r'$), both tangent at $A$ to line $AT$. Of course the dihedral angle $\theta=\angle OAO'$ between the planes of the circles is also known.
The lines perpendicular to the circles at their centers lie on plane $OAO'$ and thus meet at some point $V$. Set $d=AV$ and note that $VOAO'$ lie on a circle with diameter $AV$, so that: $$ d={OO'\over\sin\theta}={\sqrt{r^2+r'^2-2rr'\cos\theta}\over\sin\theta}. $$ From that we can also compute $OV$, $O'V$ and locate the common vertex of the cones.
Consider now a point $T$ on the common tangent and let $TH$, $TH'$ be the other tangents issued from $T$ to the circles. Note that $TH=TH'=TA$. If $THH'V$ are on the same plane, then this would be one of the tangent planes we want to find, the other one being its reflection about plane $OAO'V$. If $C$ is the projection of $A$ on $VT$, then $HC$ and $H'C$ are also perpendicular to line $VT$. The coplanarity of $THH'V$ is implied by the collinearity of $HCH'$, which in turn can be expressed as $\angle ACH+\angle ACH'=180°$.
Setting $x=AT$ we have: $$ AH=2{OA\cdot AT\over OT}=2{rx\over\sqrt{r^2+x^2}}, \quad AH'=2{r'x\over\sqrt{r'^2+x^2}}, \quad AC={VA\cdot AT\over \sqrt{VA^2+AT^2}}={dx\over\sqrt{d^2+x^2}} $$ and $$ \sin{\angle ACH\over2}={AH/2\over AC}= {r\sqrt{d^2+x^2}\over d\sqrt{r^2+x^2}}, \quad \sin{\angle ACH'\over2}={AH'/2\over AC}= {r'\sqrt{d^2+x^2}\over d\sqrt{r'^2+x^2}}. \quad $$ But $\angle ACH+\angle ACH'=180°$ implies $\sin{\angle ACH'\over2}=\cos{\angle ACH\over2}=\sqrt{1-\sin^2{\angle ACH\over2}}$. Plugging here the above results we obtain an equation for $x$, which can be solved to: $$ x^2={r^2r'^2+rr'\sqrt{r^2r'^2+d^2(d^2-r^2-r'^2)}\over d^2-r^2-r'^2}. $$ With this result we can construct $T$, $H$, $H'$ and thus find the tangent plane $VTHH'$.