If the two circles $(x - 1)^2 + (y - 3)^2 = r^2$ ($r > 0$) and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect in two distinct points, then the number of odd positive integral values of $r$ is ____.
Step-by-Step Solution
Key Concept: Two circles intersect at two distinct points if and only if the distance between centres lies strictly between the difference and sum of their radii.
The first circle has centre $C_1(1, 3)$ and radius $r_1 = r$. The second circle $x^2 + y^2 - 8x + 2y + 8 = 0$ has centre $C_2(4, -1)$ and radius $r_2 = \sqrt{16 + 1 - 8} = 3$. For two circles to intersect at two distinct points, $|r_1 - r_2| < C_1C_2 < r_1 + r_2$. Computing $C_1C_2 = \sqrt{9 + 16} = 5$, we get $|r - 3| < 5 < r + 3$, which simplifies to $2 < r < 8$. The integer values are $r = 3, 5, 7$.
Correct Answer: 3