If common tangents of `x^(2) + y^(2) = r^(2)` and `(x^2)/16 + (y^2)/(9) = 1` forms a square, then the length of diagonal of the square is

Answer

To solve the problem, we need to find the length of the diagonal of the square formed by the common tangents of the circle \(x^2 + y^2 = r^2\) and the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).

### Step-by-Step Solution:

1. **Identify the Circle and Ellipse**:
- The equation of the circle is \(x^2 + y^2 = r^2\).
- The equation of the ellipse is \(\frac{x^2}{16} + \frac{y^2}{9} = 1\). Here, \(a^2 = 16\) and \(b^2 = 9\).

2. **Find the Semi-Axes of the Ellipse**:
- The semi-major axis \(a = 4\) (since \(a = \sqrt{16}\)).
- The semi-minor axis \(b = 3\) (since \(b = \sqrt{9}\)).

3. **Calculate the Director Circle of the Circle**:
- The radius of the director circle for the circle is given by \(R = 2r\).
- Therefore, the equation of the director circle is \(x^2 + y^2 = (2r)^2 = 4r^2\).

4. **Calculate the Director Circle of the Ellipse**:
- The radius of the director circle for the ellipse is given by \(R = \sqrt{a^2 + b^2}\).
- Here, \(a^2 + b^2 = 16 + 9 = 25\), so \(R = \sqrt{25} = 5\).
- Therefore, the equation of the director circle for the ellipse is \(x^2 + y^2 = 25\).

5. **Set the Director Circles Equal**:
- Since the common tangents form a square, the radius of the director circle of the circle must equal the radius of the director circle of the ellipse.
- Thus, we set \(4r^2 = 25\).

6. **Solve for \(r\)**:
- From \(4r^2 = 25\), we get \(r^2 = \frac{25}{4}\).
- Taking the square root, we find \(r = \frac{5}{2}\).

7. **Calculate the Length of the Diagonal of the Square**:
- The diagonal \(d\) of the square formed by the common tangents is given by \(d = 2R\), where \(R\) is the radius of the director circle.
- Here, \(R = 5\), so \(d = 2 \times 5 = 10\).

### Final Answer:
The length of the diagonal of the square is \(10\) units.