Question:

<p><strong>Step 1: Set up the given information.</strong></p><p>Given circle: $x^2 + (y-1)^2 = 1$ has center $O(0, 1)$ and radius $r_1 = 1$.</p><p>Let circle C have center $P(h, k)$ and radius $r$.</p><p><strong>Step 2: Apply the external tangency condition.</strong></p><p>Since C touches the given circle externally: $|OP| = r_1 + r$</p><p>$$\sqrt{h^2 + (k-1)^2} = 1 + r \quad \text{...(1)}$$</p><p><strong>Step 3: Apply the x-axis tangency condition.</strong></p><p>Since C touches the x-axis, the distance from center to x-axis equals the radius.</p><p>For $k \geq 0$: $k = r$ (circle above x-axis)</p><p>For $k \leq 0$: The center lies on or below the x-axis (degenerate case)</p><p><strong>Step 4: Case 1 - Circle above x-axis ($k > 0$).</strong></p><p>Substitute $r = k$ into equation (1):</p><p>$$\sqrt{h^2 + (k-1)^2} = 1 + k$$</p><p>Square both sides:</p><p>$$h^2 + (k-1)^2 = (1+k)^2$$</p><p>$$h^2 + k^2 - 2k + 1 = 1 + 2k + k^2$$</p><p>$$h^2 = 4k$$</p><p>$$x^2 = 4y \quad \text{(for } y > 0 \text{)}$$</p><p><strong>Step 5: Case 2 - Circle on/below x-axis ($k \leq 0$).</strong></p><p>When the center lies on the x-axis or below, the circle with center on the negative y-axis and any radius will automatically touch the x-axis. Any point $(0, y)$ with $y \leq 0$ can be a center of such a circle.</p><p>This gives: $\{(0, y): y \leq 0\}$</p><p><strong>Step 6: Combine both cases.</strong></p><p>The complete locus is:</p><p>$$\{(x, y): x^2 = 4y\} \cup \{(0, y): y \leq 0\}$$</p><p><strong>∴ Answer:</strong> d</p>