Circles
Family of circles tangent to a line
Grade 11
Question:
<p>Consider a family of circles which are passing through the point (−1, 1) and are tangent to x-axis. If (<em>h</em>, <em>k</em>) are the co-ordinates of the centre of the circles, then the set of values of <em>k</em> is given by the interval:</p>
<p>\(0 < k < \dfrac{1}{2}\)</p>
<p>\(k \geq \dfrac{1}{2}\)</p>
<p>\(-\dfrac{1}{2} \leq k \leq \dfrac{1}{2}\)</p>
<p>\(k \leq \dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: If a circle passes through point (-1, 1) and is tangent to the x-axis, then the radius equals k (distance from center to x-axis), and the distance from center (h, k) to (-1, 1) must equal this radius. This gives us (h+1)² + (k-1)² = k².
<p><strong>Step 1:</strong> For a circle with center (h, k) tangent to the x-axis, the radius r = |k|. Since the circle passes through (-1, 1), the distance from center to this point equals the radius.</p><p><strong>Step 2:</strong> Distance formula gives: √[(h+1)² + (k-1)²] = |k|</p><p><strong>Step 3:</strong> Squaring both sides: (h+1)² + (k-1)² = k²</p><p><strong>Step 4:</strong> Expanding: (h+1)² + k² - 2k + 1 = k²</p><p><strong>Step 5:</strong> Simplifying: (h+1)² = 2k - 1</p><p><strong>Step 6:</strong> Since (h+1)² ≥ 0 for all real h, we need: 2k - 1 ≥ 0, which gives k ≥ 1/2</p><p><strong>Step 7:</strong> For any k ≥ 1/2, we can find real values of h satisfying (h+1)² = 2k - 1, confirming k ∈ [1/2, ∞)</p><p>∴ Answer: B</p>
Correct Answer: B