Consider one ellipse and one circle given by the parametric equations:
For the ellipse $$ x = x_0 + a\cdot \cos(t_1)$$ $$ y = y_0 + b\cdot \sin(t_1)$$
and for the circle
$$ x = x_1 + \rho\cdot\cos(t_2) $$ $$ y = y_1 + \rho\cdot\sin(t_2) $$
How do I find the value of $t_1$ and $t_2$ for the two tangent points of their common tangent at the up side of the curves? See this image.
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If $(x_1(t_1),y_1(t_1))$ and $(x_2(t_2),y_2(t_2))$ are the tangent points on each curve, then the slope of the common tangent can be expressed in 3 ways
$$ \frac{y_1'(t_1)}{x_1'(t_1)} = \frac{y_2'(t_2)}{x_2'(t_2)} = \frac{y_2(t_2)-y_1(t_1)}{x_2(t_2)-x_1(t_1)} $$
This gives us a system of equation that can be solved for $(t_1,t_2)$. You will get both tangents simultaneously. Depending on the geometry, you can eliminate one of them.
We need the normals to the curves. They are vectors radiating away from the curves (ellipse or circle) such that they are always at a right angle to the curve. If normals to the curves are parallel, then your tangent line will intersect each norm (one on the ellipse and one on the circle) at a right angle. We know that the desired tangent line will intersect norms that are equal on the two curves.
Take the derivative of each curve to get the tangent. It will be the tangent at all points. Circle: $$ \left[\begin{array}{c}x\\y\end{array}\right]=\left(\begin{array}{c}x_{1}\\y_{1}\end{array}\right)+\rho\left(\begin{array}{c}cos(u)\\sin(u)\end{array}\right)$$
Tangent: $$\left[\begin{array}{c}x'\\y'\end{array}\right]=\left(\begin{array}{c}-\rho\cdot sin(u)\\\rho\cdot cos(u)\end{array}\right)$$ To get the norm, swap vector positions with the tangent elements and then change the sign of one of them. It would be nice to be consistent about this. I have set the vectors to go away from the curves.
$$norm=\left(\begin{array}{c}\rho\cdot cos(u)\\\rho\cdot sin(u)\end{array}\right)$$
Similarly, the norm of the ellipse will be $$norm=\left(\begin{array}{c}b\cdot cos(t)\\a\cdot sin(t)\end{array}\right)$$ I have changed to using $t$ and $u$ for the parameters instead of $t_1$ and $t_2$. FYI to create a slope from a direction vector, use $y/x$.
Now that we have some preliminaries out of the way, here is the general idea for solution. My unknowns are point $C(x,y)$ and $D(x,y)$ and parameter $u$ and parameter $t$. Therefore there are 6 unknowns. Take a look at the picture to understand where things are.
Equation 1: The y part of the parameterized circle. $$D_y=P_{1y}+\rho\cdot sin(u)\qquad\text{eq:01}$$
Equation 2: The x part of the parameterized circle. $$D_x=P_{1x}+\rho\cdot cos(u)\qquad\text{eq:02}$$
Equation 3: The y part of the parameterized ellipse. $$C_y=P_{y}+b\cdot sin(t)\qquad\text{eq:03}$$
Equation 4: The x part of the parameterized ellipse. $$C_x=P_{x}+a\cdot cos(t)\qquad\text{eq:04}$$
Equaton 5: The slope of the two vectors in the picture have to be equal. Call $m$ the slope. $$m=\frac{a\cdot sin(t)}{b\cdot cos(t)}=\frac{\rho\cdot sin(u)}{\rho\cdot cos(u)}\qquad\text{eq:05}$$
Now if there is a tricky part, then this is it. We will write the equation of a line (in point:slope form) that has slope $-1/m$ and goes through point $C$. Then we will write the line equation that also has slope $-1/m$ and goes through point $D$. They are the same line, so we will equate them and lose the x and y to get the 6th equation.
$$y-C_y=-\frac{1}{m}\cdot (x-C_x)\qquad\text{1st line}$$ $$y-D_y=-\frac{1}{m}\cdot (x-D_x)\qquad\text{2nd line}$$
$$-\frac{1}{m}(x-C_x)-y+C_y=-\frac{1}{m}(x-D_x)-y+D_y\qquad\text{equated lines}$$ Equation 6: After some minor algebra: $$\frac{C_x}{m}+C_y=\frac{D_x}{m}+D_y\qquad\text{eq:06}$$
I am not going to solve these symbolically! You can if you have to.
Let $a=0.2\;b=0.5\;\rho=0.5\;\;P=(1,1.75)\;P_1=(2,2)$ That is shown in the picture. There are actually 4 tangent lines, 2 on each side of the circle and ellipse and 2 that cross over one another on the interior.
Just for the ones in the picture, point $C=(0.98,2.25)$ and $D=(1.87,2.48)$ $t=1.68$ and $u=1.83$