The circle passing through $(1, -2)$ and touching the $x$-axis at $(3, 0)$ also passes through the point:
Step-by-Step Solution
Key Concept: Circle touching $x$-axis at $(3, 0)$ has centre $(3, a)$ and radius $|a|$. Substituting $(1, -2)$: $(1 - 3)^2+(-2-a)^2 = a^2$ gives $a = -2$. Centre $(3, -2)$, radius $2$; check $(5, -2)$: $(5-3)^2+0 = 4$.
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Correct Answer: (2)