Let the ellipse be:
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
Set $P(a,t)$, the tangent line $PT_1$ is $y=k(x-a)+t$, $k=\tan A_1$, since it is tangent line, so put in ellipse equation, we have:
$$ b^2x^2+a^2(kx-ka+t)^2=a^2b^2$$
i.e.,
$$(b^2+a^2k^2)x^2+2a^2k(t-ka)x+a^2(t-ka)^2-a^2b^2=0$$
Solving for the point of tangency, the discriminant $\Delta$ must be zero:
$$\Delta$=$4a^4k^2(t-ka)^2-4(b^2+a^2k^2)[a^2(t-ka)^2-a^2b^2] $$
At $\Delta=0$, we get $ 2kta-t^2+b^2=0$, that is, $k=\dfrac{t^2-b^2}{2ta}$.
Let $k_1$ be $PF_1$'s slope, $k_1=\dfrac{t}{a+c}=\tan A_2$, the $ \angle T_1PF_1=|A_1-A_2|$, now we calculate $\tan(A_1-A_2)$:
$$\tan(A_1-A_2)=\dfrac{\tan A_1-\tan A_2}{1+\tan A_1 \tan A_2}=\dfrac{k-k_1}{1+k k_1}=\dfrac{\dfrac{t^2-b^2}{2ta}-\dfrac{t}{a+c}}{1+\dfrac{t^2-b^2}{2ta}\dfrac{t}{a+c}}=\dfrac{(t^2-b^2)(a+c)-2at^2}{2at^2+2act+t^3-tb^2}$$
Since $b^2=a^2-c^2$, we get:
$$ RHS=\dfrac{(t^2-a^2+c^2)(a+c)-2at^2}{t(2a^2+2ac+t^2-a^2+c^2)}=\dfrac{c-a}{t}$$
Let $k_2$ be line $PF_2$'s slope, then $k_2=\dfrac{t}{a-c}=\tan\angle PF_2T_2$. Since $\angle T_2PF_2=\dfrac{\pi}{2}-\angle PF_2T_2$, so $\tan \angle T_2PF_2=\dfrac{1}{\tan\angle PF_2T_2}=\dfrac{a-c}{t}=\tan(A_2-A_1)$. Since the angles are acute, we have:
$\angle T_2PF_2=A_2-A_1=\angle T_1PF_1$
Hint Let $O$ denote the centre of the ellipse.
Pick an arbitrary point $A$ on the ellipse. Draw a circle with the centre at $O$ and passing through $A$.
This circle will intersect the ellipse in 4 points* $A,B,C,D$. Show that $ABCD$ is a rectangle, with the edges parralel to the axes. Since the axes go through $)$ it is easy to construct them.
*If the circle intersects the ellipse in only two points, the two points are the intersection between the ellipse and one of the axes.
Of course, if the ellipse is degenerated (a circle) the construction above will fail, since there will be infinitely many points of intersection. But in that case, there is no axis to construct.
P.S. You don't even need the center to solve the problem. Given an ellipse, you can use the fact that the line passing through the midpoints of two parralel cords passes through the centre of the ellipse.
So given an ellipse without the centre, you start by doing the following:
Pick an arbitrary chord $EF$. Pick another point $G$ on the ellipse and draw the parralel through $G$ to $EF$. Let $H$ be the intersection between this parralel and the ellipse.
Pick another cord $IJ$ which is NOT parralel to $EF$. Draw another chord as above $KL \parallel EF$.
Now, if $M,N, P,Q$ are the midpoints of $EF, GH, IJ, KL$ then $MN \cap PQ= \{O\}$ is the centre of the ellipse.
Best Answer
This is an interesting problem if one insists on a solution constructable using purely geometric operations.
For simplicity in describing the solution, we will choose a coordinate system such that the two foci $A, B$ are $(1,0)$ and $(0,1)$ respectively. We will also assume the circle $\mathscr{C}$ we are dealing with is centered at $C = (\alpha,\beta)$ in the first quadrant and the radius of $\mathscr{C}$ is $r$.
It is known that the family of ellipses and hyperbolas having $A$ and $B$ as foci are given by:
$$\frac{x^2}{1+\lambda} + \frac{y^2}{\lambda} = 1\quad\text{ and is } \begin{cases} \text{ an ellipse, }&\text{ if } \lambda > 0\\ \text{ a hyperbola, } &\text{ if } 0 > \lambda > -1 \end{cases}\tag{*0}$$ For a point $p = (x,y)$ on this ellipse/hyperbola, the normal is in the direction $(\frac{x}{1+\lambda}, \frac{y}{\lambda})$. i.e. the normal line passing through $(x,y)$ is given by:
$$\left\{\;\left( x ( 1 + \frac{t}{1+\lambda} ),\,y ( 1 + \frac{t}{\lambda} ) \right) : t \in \mathbb{R}\;\right\}$$
If this ellipse/hyperbola is tangent to any circle centered at $C$ at point $p$, the above normal line with pass through $C = (\alpha,\beta)$ and one can show that $(x,y)$ lies on a cubic curve $\mathscr{P}$:
$$(\alpha y - \beta x )\left(x(x-\alpha) + y(y-\beta)\right) - (x-\alpha)(y-\beta) = 0\tag{*1}$$
This is the red curve depicted in above picture.
For $r$ not too large, $\mathscr{C}$ and $\mathscr{P}$ in general will intersect at 4 points. Two of them are "tangent" points to two ellipses and the other two are tangent points to two hyperbolas. All these four ellipses/hyperbolas are having $A, B$ as their foci. For these four points, one can simplify $(*1)$ and shows that they lie on a conic $\mathscr{Q}$: $$(\alpha y - \beta x)(\alpha x + \beta y - \gamma) - (x-\alpha)(y-\beta) = 0\tag{*2}$$ where $\gamma = \alpha^2 + \beta^2 - r^2$. This is the cyan curve containing $C, F, G, J, K, L$ in above picture.
Some of these points are relatively easy to construct.
A conic is determined by any five points on it. If for whatever means we get an extra point on $\mathscr{Q}$, then we can construct $\mathscr{Q}$ and its intersections with $\mathscr{C}$, e.g. the point $L$ in the above picture. These are the tangent points we want.
One can show that the conic $\mathscr{Q}$ intersects the $x$-axis at $J = (\mu,0)$ and $K = (1/\mu,0)$ where $\mu$ is a root of the quadratic equation: $$\alpha x^2 - (\gamma +1) x + \alpha = 0\tag{*3}$$ If one doesn't mind a little bit of algebra, we are done. Otherwise, we can convert this formula to a geometric construction of $J$ and $K$ in 3 steps:
Update Since Stefan asks, here are some details how to derive the various equations.
If the normal passes through $C = (\alpha,\beta)$, we have a $t$ such that:
$$x (1 + \frac{t}{1+\lambda}) = \alpha \quad\text{ and }\quad y (1+ \frac{t}{\lambda}) = \beta$$ Eliminating $t$ give us: $$\begin{align}(1+\lambda)(\frac{\alpha}{x}-1) = \lambda (\frac{\beta}{y} - 1 ) \iff & (1+\lambda)(x-\alpha)y - \lambda(y-\beta)\alpha = 0\\ \iff & (x-\alpha)y - \lambda( \alpha y - \beta x ) = 0 \end{align}$$ This gives us $$ \lambda = \frac{(x - \alpha)y}{\alpha y - \beta x} \quad\text{ and }\quad 1 + \lambda = \frac{(y - \beta)x}{\alpha y - \beta x} $$ Substitute this back into $(*0)$, we get $(*1)$: $$\begin{align} & (\alpha y - \beta x)\left(\frac{x}{y-\beta} + \frac{y}{x-\alpha}\right) = 1\\ \iff & (\alpha y - \beta x)\left(x(x-\alpha)+y(y-\beta)\right) - (x-\alpha)(y-\beta) = 0 \end{align}$$
If $(x,y)$ also lies on $\mathscr{C}$, the $2^{nd}$ factor in above expression can be simplified as:
$$\begin{align} x(x-\alpha)+y(y-\beta) = & (x-\alpha)^2 + (y-\beta)^2 + \alpha (x-\alpha)+\beta(y-\beta)\\ = & \alpha x + \beta y - (\alpha^2 + \beta^2 - r^2)\\ = & \alpha x + \beta y - \gamma \end{align}$$
Substituting this into $(*1)$ gives us the equation of $\mathscr{Q}$ in $(*2)$:
$$(\alpha y - \beta x)(\alpha x + \beta y - \gamma) - (x-\alpha)(y-\beta) = 0$$
Now $(*3)$ is really the equation of intersecting $\mathscr{Q}$ with the $x$-axis. We just set $y = 0$ in $(*2)$, eliminate the common factor $\beta$ and we are done.