Find the complex number z that satisfies the conditions…
Find the complex number $z$ that satisfies the conditions: $$ \left|\frac{z-12}{z-8i}\right| = \frac53, \qquad \left|\frac{z-4}{z-8}\right| = 1. $$
Explanation of Method
The first condition says the distance from $z$ to $12$ is $\frac53$ times the distance from $z$ to $8i$ — an Apollonius circle. The second says $z$ is equidistant from $4$ and $8$ — the perpendicular bisector of that segment, a vertical line. Converting both to Cartesian coordinates and intersecting the line with the circle gives the solutions.
Condition 1 · $\left|\dfrac{z-12}{z-8i}\right|=\dfrac53$
Step 1 · Cross-multiply
$$ 3|z-12| = 5|z-8i| $$
Step 2 · Substitute $z=x+iy$
$$ 3\sqrt{(x-12)^2+y^2} = 5\sqrt{x^2+(y-8)^2} $$
Squaring: $9[(x-12)^2+y^2] = 25[x^2+(y-8)^2]$.
Step 3 · Expand and simplify
$$ 9x^2-216x+1296+9y^2 = 25x^2+25y^2-400y+1600 $$
$$ -16x^2-216x-16y^2+400y-304=0 \quad\Longrightarrow\quad x^2+\frac{27}{2}x+y^2-25y+19=0 $$
(dividing by $-16$).
Step 4 · Complete the square
$$ \left(x+\frac{27}{4}\right)^2-\frac{729}{16} + \left(y-\frac{25}{2}\right)^2-\frac{625}{4}+19=0 $$
$$ \boxed{\left(x+\frac{27}{4}\right)^2+\left(y-\frac{25}{2}\right)^2 = \frac{2925}{16}} $$
An Apollonius circle with center $\left(-\frac{27}{4},\frac{25}{2}\right)$ and radius $\dfrac{15\sqrt{13}}{4}$.
Condition 2 · $\left|\dfrac{z-4}{z-8}\right|=1$
Step 5 · Equidistance condition
$$ |z-4|=|z-8| $$
the perpendicular bisector of the segment from $4$ to $8$.
Step 6 · Solve for the line
$$ (x-4)^2+y^2 = (x-8)^2+y^2 \quad\Longrightarrow\quad -8x+16=-16x+64 \quad\Longrightarrow\quad \boxed{x=6} $$
Intersect the two loci
Step 7 · Substitute $x=6$ into the circle
$$ \left(6+\frac{27}{4}\right)^2+\left(y-\frac{25}{2}\right)^2 = \frac{2925}{16} \quad\Longrightarrow\quad \left(\frac{51}{4}\right)^2+\left(y-\frac{25}{2}\right)^2=\frac{2925}{16} $$
$$ \frac{2601}{16}+\left(y-\frac{25}{2}\right)^2 = \frac{2925}{16} \quad\Longrightarrow\quad \left(y-\frac{25}{2}\right)^2 = \frac{324}{16}=\frac{81}{4} $$
Step 8 · Solve for $y$
$$ y-\frac{25}{2}=\pm\frac92 \quad\Longrightarrow\quad y=17 \ \text{ or } \ y=8 $$
Schematic (not to scale): the line $x=6$ meets the Apollonius circle at two points, $z_1=6+17i$ and $z_2=6+8i$.
Step 9 · The two solutions
$$ \boxed{z_1 = 6+17i} \qquad\qquad \boxed{z_2 = 6+8i} $$
Verification of $z_1=6+17i$
$$ |z_1-12|=|-6+17i|=\sqrt{325}=5\sqrt{13}, \qquad |z_1-8i|=|6+9i|=\sqrt{117}=3\sqrt{13} $$
$$ \frac{|z_1-12|}{|z_1-8i|}=\frac{5\sqrt{13}}{3\sqrt{13}}=\frac53 \quad \checkmark \qquad\qquad |z_1-4|=|z_1-8|=\sqrt{293} \quad \checkmark $$
Verification of $z_2=6+8i$
$$ |z_2-12|=|-6+8i|=10, \qquad |z_2-8i|=|6|=6 \quad\Longrightarrow\quad \frac{10}{6}=\frac53 \quad \checkmark $$
$$ |z_2-4|=|z_2-8|=2\sqrt{17} \quad \checkmark $$
| Locus | Equation |
|---|---|
| Condition 1 (Apollonius circle) | $\left(x+\frac{27}{4}\right)^2+\left(y-\frac{25}{2}\right)^2=\frac{2925}{16}$ |
| Condition 2 (perpendicular bisector) | $x=6$ |
Result
Both intersection points of the Apollonius circle with the vertical line $x=6$ satisfy the original system.