<p>If the two circles \((x-1)^2 + (y-3)^2 = r^2\) and \(x^2 + y^2 - 8x + 2y + 8 = 0\) intersect in two distinct points, then</p>
Step-by-Step Solution
Key Concept: Two circles intersect in two distinct points when the distance between their centers lies strictly between the difference and sum of their radii: |r₁ - r₂| < d < r₁ + r₂. You must rewrite the second circle in standard form to identify its center and radius.
<p><strong>Step 1:</strong> Write the first circle in standard form: $(x-1)^2 + (y-3)^2 = r^2$ has center $C_1 = (1, 3)$ and radius $r_1 = r$.</p><p><strong>Step 2:</strong> Rewrite the second circle by completing the square: $x^2 + y^2 - 8x + 2y + 8 = 0$ becomes $(x-4)^2 + (y+1)^2 = 16 + 1 - 8 = 9$. So center $C_2 = (4, -1)$ and radius $r_2 = 3$.</p><p><strong>Step 3:</strong> Find the distance between centers: $d = \sqrt{(4-1)^2 + (-1-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.</p><p><strong>Step 4:</strong> For two distinct intersection points, apply the condition: $|r - 3| < 5 < r + 3$.</p><p><strong>Step 5:</strong> From $5 < r + 3$, we get $r > 2$. From $|r - 3| < 5$: either $r - 3 < 5$ (giving $r < 8$) or $3 - r < 5$ (giving $r > -2$, always true for $r > 0$). Thus $2 < r < 8$.</p><p>∴ Answer: A</p>
Correct Answer: A