Question:

**Step 1:** Determine the center and radius of the given circle. The equation of the given circle is $x^2 + y^2 - 6x + 6y + 14 = 0$. To find its center and radius, we complete the square: $(x^2 - 6x + 9) + (y^2 + 6y + 9) - 9 - 9 + 14 = 0$ $(x - 3)^2 + (y + 3)^2 = 18 - 14$ $(x - 3)^2 + (y + 3)^2 = 4$ Thus, the center of the given circle is $C_1 = (3, -3)$ and its radius is $r_1 = \sqrt{4} = 2$. **Step 2:** Define the properties of the required circle. Let the center of the required circle be $C_2 = (h, k)$ and its radius be $r$. Since the required circle touches the y-axis, the distance from its center to the y-axis is equal to its radius. The distance from $(h, k)$ to the y-axis is $|h|$. Therefore, $r = |h|$. Assuming the circle is in the right half-plane (i.e., $h > 0$), we have $r = h$. **Step 3:** Apply the condition for external tangency. The required circle touches the given circle externally. For two circles to touch externally, the distance between their centers must be equal to the sum of their radii. The distance between $C_1(3, -3)$ and $C_2(h, k)$ is $\sqrt{(h - 3)^2 + (k - (-3))^2} = \sqrt{(h - 3)^2 + (k + 3)^2}$. The sum of their radii is $r + r_1 = r + 2$. Therefore, we have the equation: $\sqrt{(h - 3)^2 + (k + 3)^2} = r + 2$ **Step 4:** Substitute $r=h$ and simplify the equation. Substitute $r = h$ into the equation from Step 3: $\sqrt{(h - 3)^2 + (k + 3)^2} = h + 2$ Square both sides of the equation: $(h - 3)^2 + (k + 3)^2 = (h + 2)^2$ Expand both sides: $h^2 - 6h + 9 + (k + 3)^2 = h^2 + 4h + 4$ Isolate the term involving $k$: $(k + 3)^2 = h^2 + 4h + 4 - h^2 + 6h - 9$ $(k + 3)^2 = 10h - 5$ **Step 5:** Express the locus of the center. The equation $(k + 3)^2 = 10h - 5$ represents the locus of the center $(h, k)$ of the required circle. To express this locus in terms of $x$ and $y$, we replace $h$ with $x$ and $k$ with $y$: $(y + 3)^2 = 10x - 5$ Expand the left side: $y^2 + 6y + 9 = 10x - 5$ Rearrange the terms to form the final equation: $y^2 - 10x + 6y + 14 = 0$