Answer to Question #350108 in Analytic Geometry for Jerome

Question

Find the equation of the locus of the center of the circle which moves so that it is tangent to the y-axis and to the circle of radius one (1) with center at (2,0).

Answer 1

Let $P(x,y)$ be the center of a moving circle tangent to the y-axis and to a circle with a radius $1$ with center at $(2,0).$

The distance from point $P$ to the y-axis is $|x|$. The distance from the point $P$ to a center $(2,0)$ is

$$\sqrt{(x-2)^2+(y-0)^2}$$

 The distance from the point $P$ to a circle is 

$$\sqrt{(x-2)^2+(y-0)^2}-1$$

Then

$$\sqrt{(x-2)^2+(y-0)^2}-1=|x|$$ $$\sqrt{(x-2)^2+(y-0)^2}=|x|+1$$ $$x^2-4x+4+y^2=x^2+2|x|+1$$ $$y^2=4x+2|x|-3$$

Since $y^2\geq0, y\in \R,$ we take $|x|=x, x\geq0$

$$y^2=4x+2x-3$$ $$y^2=6x-3$$

The equation of the locus of the center of a moving circle tangent to the y-axis and to a circle with a radius 1 with center at $(2,0)$ is the equaion of the parabola

$$y^2=6x-3$$