Question:

<p><strong>Step 1:</strong> For a circle \(x^2 + y^2 + 2gx + 2fy + c = 0\), the normal at point \((x_0, y_0)\) passes through the center and the given point.</p><p><strong>Step 2:</strong> Rewrite the circle: \(x^2 + y^2 - 5x + 2y - 48 = 0\). The center is \(\left(\frac{5}{2}, -1\right)\).</p><p><strong>Step 3:</strong> The slope of the line joining center \(\left(\frac{5}{2}, -1\right)\) and point \((5, 6)\) is \(m = \frac{6-(-1)}{5-\frac{5}{2}} = \frac{7}{\frac{5}{2}} = \frac{14}{5}\).</p><p><strong>Step 4:</strong> Equation of normal: \(\frac{x-5}{y-6} = \frac{5}{7}\) gives \(7(x-5) = 5(y-6)\), so \(7x - 35 = 5y - 30\), thus \(14x - 5y - 40 = 0\).</p><p>∴ Answer is (c).</p>