If `z = 5 + t + isqrt(25 - t^(2)), (-5 le t le 5)`, then locus of z is a curve which passes through

Options

A. 5 + 0i
B. `-2 + 3i`
C. 2 + 4i
D. `-2-3i`

Answer

To find the locus of the complex number \( z = 5 + t + i\sqrt{25 - t^2} \) where \( -5 \leq t \leq 5 \), we will follow these steps:

### Step 1: Identify the real and imaginary parts of \( z \)
The complex number \( z \) can be expressed as:
\[ z = x + iy \]
where:
- \( x = 5 + t \)
- \( y = \sqrt{25 - t^2} \)

### Step 2: Express \( t \) in terms of \( x \)
From the equation for \( x \):
\[ t = x - 5 \]

### Step 3: Substitute \( t \) into the equation for \( y \)
Substituting \( t = x - 5 \) into the equation for \( y \):
\[ y = \sqrt{25 - (x - 5)^2} \]

### Step 4: Simplify the expression for \( y \)
Now, simplify \( y \):
\[ y = \sqrt{25 - (x^2 - 10x + 25)} = \sqrt{25 - x^2 + 10x - 25} = \sqrt{10x - x^2} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ y^2 = 10x - x^2 \]

### Step 6: Rearrange the equation
Rearranging the equation leads to:
\[ x^2 - 10x + y^2 = 0 \]

### Step 7: Identify the type of curve
This equation represents a circle. To find the center and radius, we can complete the square:
\[ (x - 5)^2 + y^2 = 25 \]
This indicates a circle centered at \( (5, 0) \) with a radius of \( 5 \).

### Step 8: Determine the locus of \( z \)
The locus of \( z \) is therefore a circle with center \( (5, 0) \) and radius \( 5 \).

### Step 9: Identify points through which the curve passes
The circle passes through the following points:
1. \( (5, 5) \) when \( t = 0 \)
2. \( (5, 0) \) when \( t = 5 \)
3. \( (5, -5) \) when \( t = -5 \)

### Final Result
Thus, the locus of \( z \) is a circle centered at \( (5, 0) \) with radius \( 5 \).

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