Key Concepts and Formulas

• Distance between centers of externally tangent circles: If two circles with radii $r_1$ and $r_2$ are tangent externally, the distance between their centers is $r_1 + r_2$.
• Equation of a circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
• Distance from a point to the y-axis: The distance from a point $(h, k)$ to the y-axis is $|h|$.

Step-by-Step Solution

Step 1: Define the variable circle and its properties.

Let the center of the variable circle be $(h, k)$ and its radius be $r$. Since the circle touches the y-axis, its radius is equal to the absolute value of the x-coordinate of its center, i.e., $r = |h|$. Since the circle lies in the first quadrant, $h > 0$ and $k > 0$. Thus, $r = h$.

Step 2: Use the condition that the variable circle touches the given circle externally.

The given circle has the equation $x^2 + y^2 = 1$, which means its center is at $(0, 0)$ and its radius is $1$. The variable circle touches this circle externally. Therefore, the distance between their centers is the sum of their radii:

$\sqrt{(h - 0)^2 + (k - 0)^2} = h + 1$

This equation represents the distance between the centers $(h, k)$ and $(0, 0)$ being equal to the sum of the radius of variable circle $h$ and the radius of given circle $1$.

Step 3: Simplify the equation and eliminate $h$.

Square both sides of the equation to get rid of the square root:

$h^2 + k^2 = (h + 1)^2$ $h^2 + k^2 = h^2 + 2h + 1$ $k^2 = 2h + 1$

Step 4: Express $h$ in terms of $k$.

Solve for $h$ in terms of $k$:

$2h = k^2 - 1$ $h = \frac{k^2 - 1}{2}$

Step 5: Find the locus by replacing $(h, k)$ with $(x, y)$.

To find the locus, replace $h$ with $x$ and $k$ with $y$:

$x = \frac{y^2 - 1}{2}$

Step 6: Solve for $y$.

Multiply both sides by 2:

$2x = y^2 - 1$ $y^2 = 2x + 1$ $y = \pm\sqrt{2x + 1}$

Step 7: Consider the quadrant condition.

Since the circle lies in the first quadrant, $y > 0$. Therefore, we only consider the positive square root:

$y = \sqrt{2x + 1}$

Also, since the radius $r = h > 0$, we have $x > 0$.

Step 8: State the final locus equation and condition.

The locus of the centers of the circles is $y = \sqrt{1 + 2x}, x \ge 0$

Common Mistakes & Tips

• Forgetting the quadrant condition: Make sure to consider that the circle lies in the first quadrant, which implies $h > 0$ and $k > 0$, and consequently, $x > 0$ and $y > 0$.
• Incorrectly applying the distance formula: Ensure the distance formula is applied correctly between the centers of the circles.
• Squaring the equation prematurely: While squaring eliminates the square root, it's essential to ensure no extraneous solutions are introduced. In this case, the quadrant condition helps eliminate the negative square root.

Summary

We started by defining the variable circle's center and radius. We used the condition that it touches both the y-axis and the given circle externally to establish a relationship between the coordinates of the center. We then simplified the equation, solved for one variable in terms of the other, and replaced $(h, k)$ with $(x, y)$ to find the locus. Finally, we considered the quadrant condition to arrive at the final equation of the locus.

The final answer is \boxed{x = \sqrt {1 + 2y} ,y \ge 0}, which corresponds to option (A). The final answer is \boxed{A}.