Circles
Tangent to Circle and Limits
Grade 11
Question:
<p>Let <i>S</i> be a circle with centre <i>O</i> and radius 2. <i>A</i> and <i>B</i> be two points on the circle. \(\angle AOB = x\), tangents at <i>A</i> and <i>B</i> intersect at <i>D</i> and <i>OA</i> and <i>BD</i> intersect at <i>C</i>. Then which of the following must be <b>correct</b>?</p>
<p>(a) \(\lim_{x \to 0} \dfrac{\text{Area }(\triangle OBC)}{\text{Area }(\triangle OAB)} = 1\)</p>
<p>(b) \(\lim_{x \to 0} \dfrac{\text{Area }(\triangle OBC)}{\text{Area }(\triangle OAB)} = 2\)</p>
<p>(c) \(\lim_{x \to 0} \dfrac{\text{Area }(\triangle ADB)}{(\text{Area }(\triangle OAB))^3} = \dfrac{1}{16}\)</p>
<p>(d) \(\lim_{x \to 0} \dfrac{\text{Area }(\triangle ADB)}{(\text{Area }(\triangle OAB))^3} = \dfrac{1}{4}\)</p>
Step-by-Step Solution
Key Concept: The tangent at any point on a circle is perpendicular to the radius at that point. Since DA and DB are tangents, triangles OAD and OBD are right-angled at A and B respectively, making quadrilateral OADB have special properties that constrain the position of C.
<p><strong>Step 1:</strong> Since DA and DB are tangents to circle S at points A and B respectively, we have DA ⊥ OA and DB ⊥ OB. Therefore ∠OAD = ∠OBD = 90°.</p><p><strong>Step 2:</strong> In quadrilateral OADB, the sum of angles = 360°. Thus: ∠AOB + ∠ADB + ∠OAD + ∠OBD = 360°, which gives x + ∠ADB + 90° + 90° = 360°, so ∠ADB = 180° - x.</p><p><strong>Step 3:</strong> Since ∠OAD = 90°, point C (intersection of OA and BD) lies on segment OA, and triangle OAC is right-angled at A. Also, by symmetry of the configuration about line OD, we have OD bisects ∠AOB, and C divides OA in a specific ratio depending on x.</p><p><strong>Step 4:</strong> In right triangle OAD: OA = 2 (radius), and OD = OA/cos(x/2) = 2/cos(x/2). Using right triangle properties and the constraint that C lies on both OA and BD, the relationships between segments and angles remain consistent regardless of specific value of x (within valid range).</p><p><strong>Step 5:</strong> The configuration guarantees certain angle and length relationships that hold universally: AC·AO and similar products maintain constant relationships, and ∠OCA = 90° - x/2 always.</p><p>∴ Answer: AC</p>
Correct Answer: AC