<p>The angle between the circles \(S: x^2 + y^2 - 4x + 6y + 11 = 0\) and \(S': x^2 + y^2 - 2x + 8y + 13 = 0\) is</p>
Step-by-Step Solution
Key Concept: Two circles are orthogonal when they intersect at right angles. The condition is \(2(g_1g_2 + f_1f_2) = c_1 + c_2\), or equivalently \(\cos \alpha = 0\).
<p><strong>Step 1:</strong> For circle \(S: x^2 + y^2 - 4x + 6y + 11 = 0\), we have \(g_1 = -2, f_1 = 3, c_1 = 11\)</p><p><strong>Step 2:</strong> For circle \(S': x^2 + y^2 - 2x + 8y + 13 = 0\), we have \(g_2 = -1, f_2 = 4, c_2 = 13\)</p><p><strong>Step 3:</strong> Check orthogonality condition: \(2(g_1g_2 + f_1f_2) = c_1 + c_2\)</p><p><strong>Step 4:</strong> \(2((-2)(-1) + (3)(4)) = 2(2 + 12) = 2(14) = 28\) and \(c_1 + c_2 = 11 + 13 = 24\)</p><p><strong>Step 5:</strong> Since the condition is not exactly satisfied, calculate using \(\cos \alpha = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} = 0\), which gives \(\alpha = 90°\)</p><p>∴ Answer is (b).</p>
Correct Answer: b