Circles
Locus problems
Grade 11
Question:
<p>Locus of the image of the point (2, 3) in the line \((2x - 3y + 4) + k(x - 2y + 3) = 0,\ k \in \mathbb{R}\), is a</p>
<p>straight line parallel to \(y\)-axis.</p>
<p>circle of radius \(\sqrt{2}\).</p>
<p>circle of radius \(\sqrt{3}\).</p>
<p>straight line parallel to \(x\)-axis.</p>
Step-by-Step Solution
Key Concept: The given equation represents a family of lines passing through the intersection point of 2x - 3y + 4 = 0 and x - 2y + 3 = 0. The locus of images of a fixed point under reflection across a family of concurrent lines forms a circle with the fixed point and the point of concurrency as diameter endpoints.
<p><strong>Step 1:</strong> Find the point of concurrency of the family of lines (2x - 3y + 4) + k(x - 2y + 3) = 0.</p><p>This family passes through the intersection of:</p><p>2x - 3y + 4 = 0 ... (1)</p><p>x - 2y + 3 = 0 ... (2)</p><p>From (2): x = 2y - 3</p><p>Substituting in (1): 2(2y - 3) - 3y + 4 = 0</p><p>4y - 6 - 3y + 4 = 0 → y = 2</p><p>Then x = 2(2) - 3 = 1</p><p>Fixed point of concurrency: A(1, 2)</p><p><strong>Step 2:</strong> Recognize that when a point P(2, 3) is reflected across different lines of a concurrent family, all images P' lie on a circle.</p><p>The diameter of this circle has endpoints at P(2, 3) and A(1, 2) (the fixed point).</p><p><strong>Step 3:</strong> The center of the circle is the midpoint of PA:</p><p>Center = ((2+1)/2, (3+2)/2) = (3/2, 5/2)</p><p>Radius = (1/2)√[(2-1)² + (3-2)²] = (1/2)√2 = √2/2</p><p>∴ The locus is a <strong>circle</strong>: (x - 3/2)² + (y - 5/2)² = 1/2</p>
Correct Answer: B