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

### Step 1: Rewrite the equations using \( z = x + iy \)

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

The modulus can be calculated as:
\[ |(x - 12) + iy| = \sqrt{(x - 12)^2 + y^2} \]
\[ |x + (y - 8)i| = \sqrt{x^2 + (y - 8)^2} \]

Therefore, the equation becomes:
\[ 3 \sqrt{(x - 12)^2 + y^2} = 5 \sqrt{x^2 + (y - 8)^2} \]

2. **Second equation:**
\[ |(x + iy) - 4| = |(x + iy) - 8| \]
This simplifies to:
\[ |(x - 4) + iy| = |(x - 8) + iy| \]

The modulus can be calculated as:
\[ |(x - 4) + iy| = \sqrt{(x - 4)^2 + y^2} \]
\[ |(x - 8) + iy| = \sqrt{(x - 8)^2 + y^2} \]

Therefore, the equation becomes:
\[ \sqrt{(x - 4)^2 + y^2} = \sqrt{(x - 8)^2 + y^2} \]

### Step 2: Solve the second equation

Squaring both sides of the second equation gives:
\[ (x - 4)^2 + y^2 = (x - 8)^2 + y^2 \]
Cancelling \( y^2 \) from both sides:
\[ (x - 4)^2 = (x - 8)^2 \]
Expanding both sides:
\[ x^2 - 8x + 16 = x^2 - 16x + 64 \]
Simplifying:
\[ -8x + 16 = -16x + 64 \]
Bringing all terms involving \( x \) to one side:
\[ 8x = 48 \implies x = 6 \]

### Step 3: Substitute \( x = 6 \) into the first equation

Now substitute \( x = 6 \) into the first equation:
\[ 3 \sqrt{(6 - 12)^2 + y^2} = 5 \sqrt{6^2 + (y - 8)^2} \]
This simplifies to:
\[ 3 \sqrt{36 + y^2} = 5 \sqrt{36 + (y - 8)^2} \]
Squaring both sides:
\[ 9(36 + y^2) = 25(36 + (y - 8)^2) \]
Expanding both sides:
\[ 324 + 9y^2 = 900 + 25(y^2 - 16y + 64) \]
Simplifying:
\[ 324 + 9y^2 = 900 + 25y^2 - 400y + 1600 \]
Bringing all terms to one side:
\[ 0 = 16y^2 - 400y + 2176 \]
Dividing through by 16:
\[ 0 = y^2 - 25y + 136 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -25, c = 136 \):
\[ y = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 1 \cdot 136}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 625 - 544 = 81 \]
Thus:
\[ y = \frac{25 \pm 9}{2} \]
Calculating the two possible values:
1. \( y = \frac{34}{2} = 17 \)
2. \( y = \frac{16}{2} = 8 \)

### Conclusion

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