The line `3x-2y=k` meets the circle `x^(2)+y^(2)=4r^(2)` at only one point, if `k^(2)=`

Options

A. `20 r^(2)`
B. `52 r^(2)`
C. `(52)/(9) r^(2)`
D. `(20)/(9)r^(2)`

Explanation

To solve the problem, we need to find the value of \( k^2 \) such that the line \( 3x - 2y = k \) is tangent to the circle \( x^2 + y^2 = 4r^2 \).

### Step-by-Step Solution:

1. **Identify the Circle's Properties**:
The equation of the circle is \( x^2 + y^2 = 4r^2 \).
- The center of the circle is at \( (0, 0) \).
- The radius \( R \) of the circle is \( \sqrt{4r^2} = 2r \).

2. **Equation of the Line**:
The line can be rewritten in the standard form \( 3x - 2y - k = 0 \).
Here, \( a = 3 \), \( b = -2 \), and \( c = -k \).

3. **Perpendicular Distance from the Center to the Line**:
The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is given by:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
For our case, the center of the circle is \( (0, 0) \):
\[ d = \frac{|3(0) - 2(0) - k|}{\sqrt{3^2 + (-2)^2}} = \frac{| -k |}{\sqrt{9 + 4}} = \frac{| -k |}{\sqrt{13}} = \frac{k}{\sqrt{13}} \]

4. **Set the Distance Equal to the Radius**:
Since the line is tangent to the circle, the perpendicular distance from the center to the line must equal the radius of the circle:
\[ \frac{k}{\sqrt{13}} = 2r \]

5. **Square Both Sides**:
To eliminate the square root, we square both sides:
\[ \left(\frac{k}{\sqrt{13}}\right)^2 = (2r)^2 \]
This simplifies to:
\[ \frac{k^2}{13} = 4r^2 \]

6. **Solve for \( k^2 \)**:
Multiply both sides by 13 to isolate \( k^2 \):
\[ k^2 = 4r^2 \cdot 13 = 52r^2 \]

### Final Answer:
Thus, the value of \( k^2 \) is:
\[ \boxed{52r^2} \]