<p>The common tangent to the circles \(x^2 + y^2 = 4\) and \(x^2 + y^2 + 6x + 8y - 24 = 0\) also passes through the point</p>
Step-by-Step Solution
Key Concept: For two circles touching internally, the common tangent passes through the point of contact which lies on the line joining the centres.
<p><strong>Step 1:</strong> First circle: \(x^2 + y^2 = 4\), centre \(C_1(0, 0)\) and radius \(r_1 = 2\)</p><p><strong>Step 2:</strong> Second circle: \(x^2 + y^2 + 6x + 8y - 24 = 0\), centre \(C_2(-3, -4)\) and radius \(r_2 = 7\)</p><p><strong>Step 3:</strong> Distance between centres: \(C_1C_2 = \sqrt{9 + 16} = 5\)</p><p><strong>Step 4:</strong> Since \(|C_1C_2| = 5\) and \(|r_1 - r_2| = |2 - 7| = 5\), the circles touch each other internally. The common tangent passes through \((6, -2)\).</p><p>∴ Answer is (a).</p>
Correct Answer: a