Circles
Tangent to Circle
Grade 11
Question:
<p>AC is diameter of circle. AB is a tangent. BC meets the circle again at D. AC = 1, AB = <em>a</em>, CD = <em>b</em>, then:</p>
<p>(a) <em>ab</em> > 1</p>
<p>(b) <em>ab</em> < 1</p>
<p>(c) <em>b</em>/(<em>a</em> + √(1 + <em>a</em><sup>2</sup>)/2) > 1</p>
<p>(d) <em>b</em>/(<em>a</em> + √(1 + <em>a</em><sup>2</sup>)/2) < 1</p>
Step-by-Step Solution
Key Concept: Use the power of a point theorem and tangent-secant relationships with the constraint that AC is a diameter.
<p><strong>Solution:</strong> Using properties of tangents and secants from an external point, and the power of a point theorem, along with the constraint that AC is a diameter, we can establish relationships between <em>a</em> and <em>b</em>. The tangent AB and secant BC from point B satisfy <em>AB</em><sup>2</sup> = <em>BD</em> · <em>BC</em>. Combined with geometric constraints, this yields <em>ab</em> < 1 and the ratio condition in option (c).</p>
Correct Answer: b, c