The first conclusion you have to make is that the problem is symmetric on both the $x$ and the $y$ axis, so the equation of the ellipse we want to find is of the form $$ {x^2 \over a^2}+{y^2 \over b^2}=1.$$
To find a solution to this problem we can, then, force the ellipse to be tangent to one of the circles, and the ellipse will be tangent to all the circles.
Let's focus on the circle in the first quadrant. $x_0=60$ and $y_0=25$ are the coordinates of its center and the radius is $r=5$. We can see that, if the ellipse touches the circle at $x=x_0+r$ the conic becomes a plane strip $-x_0-r \le x \le x_o+r$. In the same way, if the ellipse touches the circle at $y=y_0+r$ the conic becomes the plane strip $-y_0-r \le y \le y_0+r$.
So, let's take a point on the circle: $$x_c=x_0+r \cos \phi, y_c=y_0+r \sin \phi$$ $\phi$ has to be limited in the range $0 < \phi < {\pi \over 2}$, as we have seen that the ellipse degenerates if we pick a point on the extremes.
To force the ellipse to be tangent to the circle we will put: $$ {x_c^2 \over a^2}+{y_c^2 \over b^2}=1$$ (as the point on the circle has to be also on the ellipse).
The line tangent to the ellipse in $(x_c, y_c)$ has the equation $${x \cdot x_c \over a^2}+{y \cdot y_c \over b^2}=1 \Leftrightarrow y={b^2 \over y_c}-x{x_c \over y_c}{b^2 \over a^2}$$
The line tangent to the circle in $(x_c, y_c)$ is in the form $$y=k-{1 \over \tan \phi}x$$ so, equating the angular coefficients, we have: $${1 \over \tan \phi}={x_c \over y_c}{b^2 \over a^2}.$$
This, together with $$ {x_c^2 \over a^2}+{y_c^2 \over b^2}=1$$ gives us $$a^2=x_c^2+x_c y_c \tan\phi$$ and $$b^2=y_c^2+{x_c y_c \over \tan\phi}.$$
So we can draw an ellipse tangent to the four circles for each value of $\phi$
We have to minimize the area: that is $A=\pi a b$. To make things easier we will minimize $$f(\phi)=a^2b^2$$ we can do so as the areas are always positive, so squaring will not add minima.
Substituting everything into the equation for $f$ and differentiating in the $\phi$ variable, Wolfram Alfa gives us the nice equation: $$f'(\phi)={1 \over 64 } \csc^2 \left( {\pi \over 4}- {\phi \over 2} \right)\csc^2 \left( {\pi \over 2}+ {\phi \over 2} \right)\csc^2 \left( {\phi \over 2} \right)\sec^2 \left( {\phi \over 2} \right)\left( r+x_0 \cos \phi + y_0 \sin \phi \right)\left( -3rx_0 \sin \phi - rx_0 \sin 3 \phi -3ry_0 \cos \phi -ry_0 \cos 3 \phi -4x_0y_0 \right)\left( r \cos 2 \phi + x_0 \cos \phi - y_0 \sin \phi \right)$$
The only term that changes sign in this equation in our range of interest is the last one, so the ellipse with the minimum area (we know it is a minima as the area becomes plus infinite at the extremes of the range) has $\phi$ that is solution to $$r \cos 2 \phi + x_0 \cos \phi - y_0 \sin \phi = 0.$$ This leads to the solution of a fourth order equation, so I will not give you its closed-form solution.
The approximate value, for your numbers, is $\phi = 0.56369$ giving an area $A=13612.7$.
EDIT. It seems Jens is true. I trusted too much in Wolfram Alfa and did not double check my values.
Anyway the equation to solve was the right one and its solution gives the correct value of $\phi = 1.127$ thus implying $A=11525.3$.
In the graph a plot of the derivative (only the sign changing part) and of the area (scaled by 100) is shown. Just to show that the minima and the zeros are aligned.
