The number of possible integral values of $m$ for which the circle $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y + m^2 = 0$ have exactly two common tangents is __________.
Step-by-Step Solution
Key Concept: Two circles have exactly two common tangents when they intersect at two points, requiring $|r_1 - r_2| < AB < r_1 + r_2$.
Circle (i): $x^2 + y^2 = 4$ has centre $A(0,0)$ and radius $r_1 = 2$. Circle (ii): $x^2 + y^2 - 6x - 8y + m^2 = 0$ has centre $B(3, 4)$ and radius $r_2 = \sqrt{25 - m^2}$. For exactly two common tangents: $|r_1 - r_2| 24$ from condition (ii). Combining: $-4 < m < 4$. The integral values are $m = -3, -2, -1, 0, 1, 2, 3$, yielding 7 possible values.
Correct Answer: 7