A circle $C$ passes through the point $(4, 0)$ and touches the circle $x^2 + y^2 + 4x - 6y = 12$ externally at $(1, -1)$. The radius of $C$ is:
Step-by-Step Solution
Key Concept: The tangent to the given circle at $(1, -1)$ is also tangent to $C$ there. Write this tangent line, then find the centre of $C$ on the normal through $(1, -1)$ such that $C$ passes through $(4, 0)$. Distance from centre $(4, -5)$ to $(1, -1)$ gives radius $5$.
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Correct Answer: (1)