Circles
Locus of centre
Grade 11
Question:
<p>The locus of the centres of the circles, which touch the circle, \(x^2 + y^2 = 1\) externally, also touch the <em>y</em>-axis and lie in the first quadrant, is:</p>
<p>\(x = \sqrt{1 + 4y},\; y \geq 0\)</p>
<p>\(y = \sqrt{1 + 2x},\; x \geq 0\)</p>
<p>\(y = \sqrt{1 + 4x},\; x \geq 0\)</p>
<p>\(x = \sqrt{1 + 2y},\; y \geq 0\)</p>
Step-by-Step Solution
Key Concept: A circle touching the y-axis has its center at distance equal to its radius from the y-axis. Combined with external tangency to x² + y² = 1, this creates a constraint on the locus.
<p><strong>Step 1:</strong> Let the center of the variable circle be P(h, k) with radius r, where h > 0, k > 0 (first quadrant).</p><p><strong>Step 2:</strong> Since the circle touches the y-axis, the distance from P to y-axis equals r:<br>r = h</p><p><strong>Step 3:</strong> Since the circle touches x² + y² = 1 externally, the distance between centers equals sum of radii:<br>√(h² + k²) = 1 + r = 1 + h</p><p><strong>Step 4:</strong> Square both sides:<br>h² + k² = (1 + h)²<br>h² + k² = 1 + 2h + h²<br>k² = 1 + 2h</p><p><strong>Step 5:</strong> Replace (h, k) with (x, y) for the locus:<br>y² = 1 + 2x (or equivalently y² = 2x + 1)</p><p>This is a parabola with vertex at (-1/2, 0), opening rightward, valid for x > 0.</p><p>∴ Answer: C</p>
Correct Answer: C