<p>System of equations |<em>z</em> + 3| − |<em>z</em> − 3| = 6 and |<em>z</em> − 4| = <em>r</em>, where <em>r</em> ∈ ℝ⁺ has</p>
<p>(1) one solution if <em>r</em> > 1</p>
<p>(2) one solution if <em>r</em> < 1</p>
<p>(3) two solutions if <em>r</em> = 1</p>
<p>(4) at least one solution</p>
Step-by-Step Solution
Key Concept: The equation |z + 3| − |z − 3| = 6 defines a ray (part of a hyperbola branch) by the difference-of-distances property. The circle |z − 4| = r intersects this ray, and we need to find the critical value of r for exactly one solution.
<p><strong>Step 1:</strong> Recognize |z + 3| − |z − 3| = 6 as the locus of points where the difference of distances from F₁ = −3 and F₂ = 3 equals 6. Since 2a = 6, we have a = 3, and 2c = 6, so c = 3. This gives b² = c² − a² = 0.</p><p><strong>Step 2:</strong> When b = 0, the hyperbola degenerates into two rays on the real axis. Specifically, |z + 3| − |z − 3| = 6 holds only for x ≤ −3 (the left ray), since distance to −3 minus distance to 3 equals 6 only when z is far left on the real axis.</p><p><strong>Step 3:</strong> So we need the circle |z − 4| = r to intersect the ray {z = x + 0i : x ≤ −3}. The center of the circle is at 4 on the real axis. The closest point on the ray to the center 4 is z = −3, with distance |4 − (−3)| = 7.</p><p><strong>Step 4:</strong> For exactly one intersection point (tangency), we need r = 7. For 0 < r < 7, there are zero intersections; for r > 7, there are two intersections.</p><p>∴ Answer: 4</p>
Correct Answer: 4