If z is a complex number which simultaneously satisfies the equations
`3abs(z-12)=5abs(z-8i) " and " abs(z-4) =abs(z-8)`, where `i=sqrt(-1)`, then Im(z) can be

Options

A. 8
B. 17
C. 7
D. 15

Answer

To solve the problem, we need to find the imaginary part of the complex number \( z \) that satisfies the given equations:

1. \( 3 |z - 12| = 5 |z - 8i| \)
2. \( |z - 4| = |z - 8| \)

Let’s denote \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part of the complex number.

### Step 1: Rewrite the equations

**Equation 1:**
\[ 3 |(x + iy) - 12| = 5 |(x + iy) - 8i| \]
This simplifies to:
\[ 3 |(x - 12) + iy| = 5 |x + i(y - 8)| \]

**Equation 2:**
\[ |z - 4| = |z - 8| \]
This simplifies to:
\[ |(x + iy) - 4| = |(x + iy) - 8| \]

### Step 2: Calculate the moduli

**For Equation 1:**
\[ |(x - 12) + iy| = \sqrt{(x - 12)^2 + y^2} \]
\[ |x + i(y - 8)| = \sqrt{x^2 + (y - 8)^2} \]
Substituting these into Equation 1 gives:
\[ 3 \sqrt{(x - 12)^2 + y^2} = 5 \sqrt{x^2 + (y - 8)^2} \]

**For Equation 2:**
\[ |(x - 4) + iy| = \sqrt{(x - 4)^2 + y^2} \]
\[ |(x - 8) + iy| = \sqrt{(x - 8)^2 + y^2} \]
Substituting these into Equation 2 gives:
\[ \sqrt{(x - 4)^2 + y^2} = \sqrt{(x - 8)^2 + y^2} \]

### Step 3: Square both sides

**For Equation 1:**
Squaring both sides:
\[ 9((x - 12)^2 + y^2) = 25(x^2 + (y - 8)^2) \]
Expanding both sides:
\[ 9(x^2 - 24x + 144 + y^2) = 25(x^2 + y^2 - 16y + 64) \]
This simplifies to:
\[ 9x^2 - 216x + 1296 + 9y^2 = 25x^2 + 25y^2 - 400y + 1600 \]
Rearranging gives:
\[ -16x^2 - 16y^2 + 400y - 216x - 304 = 0 \]

**For Equation 2:**
Squaring both sides:
\[ (x - 4)^2 + y^2 = (x - 8)^2 + y^2 \]
This simplifies to:
\[ (x - 4)^2 = (x - 8)^2 \]
Expanding gives:
\[ x^2 - 8x + 16 = x^2 - 16x + 64 \]
Rearranging gives:
\[ 8x = 48 \implies x = 6 \]

### Step 4: Substitute \( x \) back into Equation 1

Substituting \( x = 6 \) into the rearranged Equation 1:
\[ -16(6^2) - 16y^2 + 400y - 216(6) - 304 = 0 \]
Calculating gives:
\[ -576 - 16y^2 + 400y - 1296 - 304 = 0 \]
Combining terms gives:
\[ -16y^2 + 400y - 2176 = 0 \]
Dividing through by -16:
\[ y^2 - 25y + 136 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 1 \cdot 136}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 625 - 544 = 81 \]
Thus:
\[ y = \frac{25 \pm 9}{2} \]
This gives:
\[ y = \frac{34}{2} = 17 \quad \text{and} \quad y = \frac{16}{2} = 8 \]

### Conclusion

The imaginary part \( \text{Im}(z) \) can be \( 8 \) or \( 17 \).