<p>If q be the angle between two tangents which are drawn to the circle \(x^2 + y^2 - 6\sqrt{3}x - 6y + 27 = 0\) from the origin, then \(2\sqrt{3}\tan q\) equals ............</p>
Step-by-Step Solution
Key Concept: Find the centre and radius of the circle. Use the relationship sin(θ/2) = r/d where d is distance from external point to centre.
<p><strong>Solution approach:</strong> First, rewrite the circle in standard form: \(x^2 + y^2 - 6\sqrt{3}x - 6y + 27 = 0\). Completing the square: \((x - 3\sqrt{3})^2 + (y - 3)^2 = 27 + 9 - 27 = 9\). So centre is \(C(3\sqrt{3}, 3)\) and radius is \(r = 3\). Distance from origin O to centre: \(OC = \sqrt{27 + 9} = 6\). If \(\theta\) is angle between two tangents from O, then \(\sin(\theta/2) = \frac{r}{OC} = \frac{3}{6} = \frac{1}{2}\), so \(\theta/2 = 30°\) and \(\theta = 60°\). Thus \(\tan(\theta/2) = \tan 30° = \frac{1}{\sqrt{3}}\), and \(\tan\theta = \tan 60° = \sqrt{3}\). Therefore \(2\sqrt{3}\tan\theta = 2\sqrt{3} \cdot \sqrt{3} = 6\).</p>
Correct Answer: 6