The set of values of k for which the equation `zbarz+(-3+4i)barz-(3+4i)z+k=0`
represents a circle, is

Options

A. `(-infty, 25]`
B. `[25,infty)`
C. `[5,infty)`
D. `(-infty,5)`

Explanation

To solve the equation \( \overline{z}z + (-3 + 4i)\overline{z} - (3 + 4i)z + k = 0 \) and find the set of values of \( k \) for which it represents a circle, we can follow these steps:

### Step 1: Rewrite the equation in standard form
The given equation is:
\[ \overline{z}z + (-3 + 4i)\overline{z} - (3 + 4i)z + k = 0 \]
This can be rearranged as:
\[ \overline{z}z + (-3 + 4i)\overline{z} - (3 + 4i)z + k = 0 \]

### Step 2: Identify the coefficients
In the standard form of a circle in the complex plane, we have:
\[ zz^* + \alpha z^* + \alpha^* z + r = 0 \]
where \( \alpha \) is a complex number and \( r \) is a constant. Here, we can identify:
- \( \alpha = -3 + 4i \)
- \( \alpha^* = -3 - 4i \)
- The constant term is \( k \).

### Step 3: Calculate the modulus of \( \alpha \)
To find the radius of the circle, we need to calculate the modulus of \( \alpha \):
\[ |\alpha| = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 4: Set up the condition for a circle
For the equation to represent a circle, the radius must be non-negative. The radius \( r \) can be expressed as:
\[ r = |\alpha|^2 - k \]
Thus, we have:
\[ 5^2 - k \geq 0 \]
which simplifies to:
\[ 25 - k \geq 0 \]

### Step 5: Solve for \( k \)
Rearranging the inequality gives:
\[ k \leq 25 \]

### Conclusion
The set of values of \( k \) for which the equation represents a circle is:
\[ k \in (-\infty, 25] \]