Question:

<p><strong>Step 1:</strong> The equation $|z - (4 + 8i)| = 10$ represents a circle with centre $(4, 8)$ and radius $10$.</p><p><strong>Step 2:</strong> Check if this forms an ellipse:</p><p>$$|(3 + 5i) - (5 + 11i)| = |-2 - 6i| = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$$</p><p>Since $2\sqrt{10} < 4\sqrt{5}$, the equation $|z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5}$ represents an ellipse with foci $S_1(3, 5)$ and $S_2(5, 11)$.</p><p><strong>Step 3:</strong> Find the parameters of the ellipse:</p><p>Distance between foci: $2ae = 2\sqrt{10}$, so $ae = \sqrt{10}$</p><p>$2a = 4\sqrt{5}$, so $a = 2\sqrt{5}$</p><p>$e = \frac{ae}{a} = \frac{\sqrt{10}}{2\sqrt{5}} = \frac{1}{\sqrt{2}}$</p><p>$$b^2 = a^2(1 - e^2) = 20\left(1 - \frac{1}{2}\right) = 10$$</p><p><strong>Step 4:</strong> The centre of the ellipse is the midpoint of $S_1$ and $S_2$:</p><p>$$\text{Centre} = \left(\frac{3 + 5}{2}, \frac{5 + 11}{2}\right) = (4, 8)$$</p><p>The radius of the circle is $10 = \sqrt{a^2 + b^2} = \sqrt{20 + 10}$, which equals the semi-major axis of the ellipse.</p><p><strong>Step 5:</strong> Since the centres coincide and the radius of the circle equals the semi-major axis of the ellipse, the circle and ellipse intersect at exactly <strong>2</strong> points.</p><p>∴ Answer is <strong>2</strong>.</p>