Circles
Tangents from External Point
Grade 11
Question:
<p>In each of the following one or more options are correct. Choose the correct option(s).</p><p>(c) The range of values of <em>a</em> such that the angle \(\theta\) between the pair of tangents drawn from \((a, 0)\) to the circle \(x^2 + y^2 = 1\) satisfies \(\frac{\pi}{2} < \theta < \pi\), \(a > 0\) has the equation</p>
<p>A. \((1, 2)\)</p>
<p>B. \((1, \sqrt{2})\)</p>
<p>C. \((-\sqrt{2}, -1)\)</p>
<p>D. \((-\sqrt{2}, -1) \cup (1, \sqrt{2})\)</p>
Step-by-Step Solution
Key Concept: The angle θ between two tangents from external point (a,0) to circle x²+y²=1 relates to the distance 'a' via: sin(θ/2) = 1/a. The constraint π/2 < θ < π translates to finding when 1/√2 < 1/a < 1, which gives a > √2.
<p><strong>Step 1:</strong> For a point P(a, 0) external to circle x² + y² = 1, if two tangents touch the circle at points T₁ and T₂, and θ is the angle between them, then in triangle OPT₁ (where O is center): sin(θ/2) = 1/a (using geometry of tangent).</p><p><strong>Step 2:</strong> Given constraint: π/2 < θ < π</p><p>Dividing by 2: π/4 < θ/2 < π/2</p><p>Taking sine: sin(π/4) < sin(θ/2) < sin(π/2)</p><p>Therefore: 1/√2 < 1/a < 1</p><p><strong>Step 3:</strong> From 1/a < 1: we get a > 1 (point outside circle)</p><p>From 1/√2 < 1/a: we get a < √2, which means a > √2 (inverting inequality)</p><p><strong>Step 4:</strong> Combining both conditions: a > √2</p><p>∴ Answer: <strong>D (a > √2)</strong></p>
Correct Answer: D