To solve the problem, we need to find the area of the square formed by the common tangents of the given circle and ellipse. Let's break down the solution step by step.
### Step 1: Identify the equations of the curves
The first curve is a circle given by:
\[
x^2 + y^2 = r^2
\]
The second curve is an ellipse given by:
\[
\frac{x^2}{16} + \frac{y^2}{9} = 1
\]
### Step 2: Determine the director circles
The director circle of a circle \(x^2 + y^2 = r^2\) is given by:
\[
x^2 + y^2 = 2r^2
\]
For the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\), the director circle is given by:
\[
x^2 + y^2 = a^2 + b^2
\]
where \(a^2 = 16\) and \(b^2 = 9\). Thus, we have:
\[
x^2 + y^2 = 16 + 9 = 25
\]
### Step 3: Set up equations for the points on the director circles
Let the point \( (h, k) \) be a vertex of the square formed by the common tangents. This point lies on both director circles:
1. From the circle:
\[
h^2 + k^2 = 2r^2
\]
2. From the ellipse:
\[
h^2 + k^2 = 25
\]
### Step 4: Equate the two equations
From the above two equations, we can equate:
\[
2r^2 = 25 \implies r^2 = \frac{25}{2}
\]
### Step 5: Find the equations of the tangents
The equation of the tangent to the circle in slope form is:
\[
y = mx + r\sqrt{1 + m^2}
\]
For the ellipse, the equation of the tangent is:
\[
y = mx + \sqrt{16m^2 + 9}
\]
### Step 6: Equate the constant terms
Since both tangents have the same slope \(m\), we can equate their constant terms:
\[
r\sqrt{1 + m^2} = \sqrt{16m^2 + 9}
\]
Substituting \(r^2 = \frac{25}{2}\) gives:
\[
\sqrt{\frac{25}{2}}\sqrt{1 + m^2} = \sqrt{16m^2 + 9}
\]
### Step 7: Square both sides and simplify
Squaring both sides, we get:
\[
\frac{25}{2}(1 + m^2) = 16m^2 + 9
\]
Multiplying through by 2 to eliminate the fraction:
\[
25(1 + m^2) = 32m^2 + 18
\]
Expanding and rearranging gives:
\[
25 + 25m^2 = 32m^2 + 18 \implies 7m^2 = 7 \implies m^2 = 1
\]
Thus, \(m = 1\) or \(m = -1\).
### Step 8: Find the tangent equations
Using \(m = 1\):
\[
y = x + r\sqrt{2} = x + 5
\]
Using \(m = -1\):
\[
y = -x + r\sqrt{2} = -x + 5
\]
### Step 9: Find the intercepts
The intercepts of the lines \(y = x + 5\) and \(y = -x + 5\) are both at \(5\) on the axes. Thus, the square formed has vertices at:
\((5, 0)\), \((0, 5)\), \((-5, 0)\), and \((0, -5)\).
### Step 10: Calculate the area of the square
The length of each side of the square is the distance between the points \((5, 0)\) and \((0, 5)\):
\[
\text{Length} = \sqrt{(5 - 0)^2 + (0 - 5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}
\]
Thus, the area of the square is:
\[
\text{Area} = (\text{Length})^2 = (5\sqrt{2})^2 = 50 \text{ square units}
\]
### Final Answer
The area of the square is:
\[
\boxed{50}
\]
