Animated Solution for Mathematics - Circles: The locus of the centres of the circles, which touch the circle, $x^2+y^2=1$ externally, also touch the y-axis and lie in the first quadrant, is :

Unveiling the Parabolic Path

Welcome, fellow traveler in the realm of coordinate geometry! Today, we are going to unravel a beautiful problem that might seem daunting at first, but is actually a masterclass in translating geometric intuition into algebraic elegance.

We are looking for the locus of the center of a moving circle that dances around a fixed circle, always maintaining a perfect, external touch. Let's break this down step-by-step.

The Foundation

First, let's ground ourselves. We have a fixed circle, $x^2+y^2=1$. This is our anchor, centered at the origin $(0,0)$ with a radius $R=1$.

Now, we introduce our moving circle. Let its center be $(h,k)$ and its radius be $r$. Because the problem explicitly states this circle resides in the first quadrant, we know for certain that $h>0$ and $k>0$.

The Geometric Constraints

Now, let's listen to the constraints. The problem gives us two vital clues.

First, the moving circle touches the $y$-axis. If a circle touches the $y$-axis, the perpendicular distance from its center $(h,k)$ to the $y$-axis must be exactly its radius $r$.

The distance from $(h,k)$ to the $y$-axis is simply the $x$-coordinate, $h$. Therefore, we have our first breakthrough:

The Tangency Condition

Next, the moving circle touches the fixed circle externally. When two circles touch externally, the distance $d$ between their centers is the sum of their radii.

The distance between the origin $(0,0)$ and the center $(h,k)$ is given by the distance formula:

The sum of the radii is $R+r=1+r$. Equating these, we get:

Now, substitute our earlier finding $r=h$ into this equation. We get:

The Algebraic Dance

We have our equation: $\sqrt{h^2+k^2}=1+h$. To find the locus, we need to eliminate the square root by squaring both sides:

Expanding the right side gives us:

Look at that! The $h^2$ terms on both sides cancel out beautifully, leaving us with:

To find the locus, we replace $(h,k)$ with the general coordinates $(x,y)$. This yields the final equation:

$$y^2=1+2x$$

Since $y$ must be positive in the first quadrant, we take the positive root: $y=\sqrt{1+2x}$. This is the equation of a parabola, a classic and elegant result.