The number of complex numbers `z` satisfying `|z-3-i|=|z-9-i|a n d|z-3+3i|=3` are a. one b. two c. four d. none of these

Options

A. one
B. two
C. four
D. none of these

Explanation

To solve the problem, we need to find the number of complex numbers \( z \) that satisfy the following two conditions:

1. \( |z - 3 - i| = |z - 9 - i| \)
2. \( |z - 3 + 3i| = 3 \)

Let's denote \( z \) as \( x + iy \), where \( x \) and \( y \) are real numbers.

### Step 1: Analyze the first condition

The first condition can be rewritten as:

\[ |z - 3 - i| = |z - 9 - i| \]

Substituting \( z = x + iy \):

\[ | (x - 3) + i(y - 1) | = | (x - 9) + i(y - 1) | \]

This means:

\[ \sqrt{(x - 3)^2 + (y - 1)^2} = \sqrt{(x - 9)^2 + (y - 1)^2} \]

Squaring both sides:

\[ (x - 3)^2 + (y - 1)^2 = (x - 9)^2 + (y - 1)^2 \]

### Step 2: Simplify the equation

We can cancel \( (y - 1)^2 \) from both sides:

\[ (x - 3)^2 = (x - 9)^2 \]

Expanding both sides:

\[ x^2 - 6x + 9 = x^2 - 18x + 81 \]

### Step 3: Rearranging the equation

Rearranging gives:

\[ -6x + 9 = -18x + 81 \]

Adding \( 18x \) to both sides:

\[ 12x + 9 = 81 \]

Subtracting 9 from both sides:

\[ 12x = 72 \]

Dividing by 12:

\[ x = 6 \]

### Step 4: Substitute \( x \) into the second condition

Now we substitute \( x = 6 \) into the second condition:

\[ |z - 3 + 3i| = 3 \]

Substituting \( z = 6 + iy \):

\[ | (6 - 3) + i(y + 3) | = 3 \]

This simplifies to:

\[ | 3 + i(y + 3) | = 3 \]

### Step 5: Solve the modulus equation

Calculating the modulus:

\[ \sqrt{3^2 + (y + 3)^2} = 3 \]

Squaring both sides:

\[ 9 + (y + 3)^2 = 9 \]

This simplifies to:

\[ (y + 3)^2 = 0 \]

### Step 6: Solve for \( y \)

Taking the square root:

\[ y + 3 = 0 \implies y = -3 \]

### Step 7: Find the complex number \( z \)

Now we have:

\[ x = 6 \quad \text{and} \quad y = -3 \]

Thus, the complex number \( z \) is:

\[ z = 6 - 3i \]

### Conclusion

There is only one complex number \( z \) that satisfies both conditions. Therefore, the answer is:

**a. one**

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