Circles
Locus and ratio of distances
Grade 11
Question:
<p><span class="math inline">A\)</span>, <span class="math inline">B\)</span> and <span class="math inline">C\)</span> are points in the xy-plane such that <span class="math inline">A(1, 2)\)</span>; <span class="math inline">B(5, 6)\)</span> and <span class="math inline">AC = 3BC\)</span>. Then:</p>
<p>(a) ABC is a unique triangle</p>
<p>(b) There can be only two such triangles.</p>
<p>(c) No such triangle is possible</p>
<p>(d) There can be infinite number of such triangles.</p>
Step-by-Step Solution
Key Concept: The locus of points C satisfying AC = 3BC forms a circle (Apollonius circle). Since this circle contains infinitely many points, there are infinitely many possible triangles ABC.
<p><strong>Step 1: Understand the constraint.</strong> We have A(1, 2), B(5, 6), and the condition AC = 3BC. Point C must satisfy this ratio condition.</p><p><strong>Step 2: Find the locus of point C.</strong> Let C(x, y) be any point satisfying AC = 3BC.<br/>√[(x-1)² + (y-2)²] = 3√[(x-5)² + (y-6)²]</p><p><strong>Step 3: Square both sides.</strong> (x-1)² + (y-2)² = 9[(x-5)² + (y-6)²]<br/>x² - 2x + 1 + y² - 4y + 4 = 9[x² - 10x + 25 + y² - 12y + 36]<br/>x² - 2x + y² - 4y + 5 = 9x² - 90x + 9y² - 108y + 549</p><p><strong>Step 4: Simplify.</strong> 0 = 8x² - 88x + 8y² - 104y + 544<br/>0 = x² - 11x + y² - 13y + 68<br/>(x - 11/2)² + (y - 13/2)² = 121/4 + 169/4 - 68 = 290/4 - 272/4 = 18/4 = 9/2</p><p><strong>Step 5: Identify the geometric shape.</strong> This is a circle with center (11/2, 13/2) and radius √(9/2) = 3/√2. Since C lies on this circle, there are infinitely many possible positions for C.</p><p><strong>Step 6: Verify validity.</strong> The circle is well-defined (center and radius exist), and points A and B are not on the locus (they don't satisfy the equation), so non-degenerate triangles can be formed with infinitely many positions of C on this circle.</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D