Problem Analysis #14

[THOUGHT] To find the parameters of the circle $S$, I need to first translate the given equation $|x + iy - z_1| = 2|x + iy - z_2|$ into a standard circle equation.

Given:

$$z_1 = 1 + 2i$$ $$z_2 = 3i$$ $$S = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + iy - z_1| = 2|x + iy - z_2|\}$$

Let $z = x + iy$. Then the equation becomes:

$$|z - z_1| = 2|z - z_2|$$

Substitute $z_1$ and $z_2$:

$$|x + iy - (1 + 2i)| = 2|x + iy - 3i|$$ $$|(x - 1) + i(y - 2)| = 2|(x) + i(y - 3)|$$

The modulus of a complex number $a + bi$ is $\sqrt{a^2 + b^2}$. Therefore, we have:

$$\sqrt{(x - 1)^2 + (y - 2)^2} = 2\sqrt{x^2 + (y - 3)^2}$$

Square both sides to eliminate the square roots:

$$(x - 1)^2 + (y - 2)^2 = 4(x^2 + (y - 3)^2)$$

Expand both sides:

$$(x^2 - 2x + 1) + (y^2 - 4y + 4) = 4(x^2 + y^2 - 6y + 9)$$ $$x^2 - 2x + 1 + y^2 - 4y + 4 = 4x^2 + 4y^2 - 24y + 36$$

Combine like terms:

$$x^2 + y^2 - 2x - 4y + 5 = 4x^2 + 4y^2 - 24y + 36$$

Move all terms to one side:

$$x^2 + y^2 - 2x - 4y + 5 - 4x^2 - 4y^2 + 24y - 36 = 0$$ $$-3x^2 - 3y^2 - 2x + 20y - 31 = 0$$

Divide the entire equation by $-3$ to simplify:

$$x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0$$

To convert this into the standard circle equation form $(x - x_0)^2 + (y - y_0)^2 = r^2$, we need to complete the square for $x$ and $y$.

For $x$:

$$x^2 + \frac{2}{3}x = \left(x + \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 = \left(x + \frac{1}{3}\right)^2 - \frac{1}{9}$$

For $y$:

$$y^2 - \frac{20}{3}y = \left(y - \frac{10}{3}\right)^2 - \left(\frac{10}{3}\right)^2 = \left(y - \frac{10}{3}\right)^2 - \frac{100}{9}$$

Substitute back into the equation:

$$\left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \left(y - \frac{10}{3}\right)^2 - \frac{100}{9} + \frac{31}{3} = 0$$ $$\left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 - \frac{101}{9} + \frac{93}{9} = 0$$ $$\left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 - \frac{8}{9} = 0$$ $$\left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 = \frac{8}{9}$$

From this equation, we identify the center $(x_0, y_0)$ and the radius $r$:

$$x_0 = -\frac{1}{3}, \quad y_0 = \frac{10}{3}, \quad r^2 = \frac{8}{9}$$

Thus, $r = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

Now, calculate $9(x_0 + y_0)r^2$:

$$9\left(-\frac{1}{3} + \frac{10}{3}\right)\left(\frac{8}{9}\right)$$ $$9\left(\frac{9}{3}\right)\left(\frac{8}{9}\right)$$ $$9 \cdot 3 \cdot \frac{8}{9}$$ $$3 \cdot 8 = 24$$

Therefore, the value of $9(x_0 + y_0)r^2$ is:

$$\boxed{24}$$

[/THOUGHT]