`C_(1):x^(2)+y^(2)=r^(2)and C_(2):(x^(2))/(16)+(y^(2))/(9)=1` interset at four distinct points A,B,C, and D. Their common tangents form a peaallelogram A'B'C'D'.
if A'B'C'D' is a square, then r is equal to

Answer

To solve the problem, we need to find the value of \( r \) such that the common tangents to the circles \( C_1: x^2 + y^2 = r^2 \) and \( C_2: \frac{x^2}{16} + \frac{y^2}{9} = 1 \) form a square.

### Step-by-step Solution:

1. **Identify the equations of the circles**:
- Circle \( C_1 \): \( x^2 + y^2 = r^2 \)
- Circle \( C_2 \): \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \)

2. **Rewrite the equation of Circle \( C_2 \)**:
- Multiply through by 144 (the least common multiple of 16 and 9) to eliminate the denominators:
\[ 9x^2 + 16y^2 = 144 \]

3. **Eliminate \( x^2 \)** from the equations:
- From \( C_1 \), express \( x^2 \) in terms of \( y^2 \):
\[ x^2 = r^2 - y^2 \]
- Substitute this into the equation of \( C_2 \):
\[ 9(r^2 - y^2) + 16y^2 = 144 \]

4. **Simplify the equation**:
- Distributing the 9:
\[ 9r^2 - 9y^2 + 16y^2 = 144 \]
- Combine like terms:
\[ 9r^2 + 7y^2 = 144 \]

5. **Rearranging for \( y^2 \)**:
\[ 7y^2 = 144 - 9r^2 \]
\[ y^2 = \frac{144 - 9r^2}{7} \]

6. **Set the condition for a square**:
- For the points \( A, B, C, D \) to form a square, the lengths of the tangents from the center to the points must be equal, which implies \( x^2 = y^2 \).
- Thus, set \( x^2 = y^2 \):
\[ x^2 = \frac{144 - 9r^2}{7} \]
- Substitute \( x^2 \) into the equation:
\[ x^2 = r^2 - y^2 \]
- Therefore:
\[ \frac{144 - 9r^2}{7} = r^2 - \frac{144 - 9r^2}{7} \]

7. **Combine and solve for \( r^2 \)**:
- Multiply both sides by 7 to eliminate the fraction:
\[ 144 - 9r^2 = 7r^2 - (144 - 9r^2) \]
- Simplifying gives:
\[ 144 - 9r^2 = 7r^2 - 144 + 9r^2 \]
- Combine like terms:
\[ 144 = 16r^2 - 144 \]
- Rearranging gives:
\[ 288 = 16r^2 \]
- Thus:
\[ r^2 = \frac{288}{16} = 18 \]

8. **Taking the square root**:
\[ r = \sqrt{18} = 3\sqrt{2} \]

### Final Answer:
\[ r = 3\sqrt{2} \]