Question #350108

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).


1
Expert's answer
2022-06-13T01:08:13-0400

Let P(x,y)P(x,y) be the center of a moving circle tangent to the y-axis and to a circle with a radius 11 with center at (2,0).(2,0).

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


(x2)2+(y0)2\sqrt{(x-2)^2+(y-0)^2}

 The distance from the point PP to a circle is 

(x2)2+(y0)21\sqrt{(x-2)^2+(y-0)^2}-1


Then


(x2)2+(y0)21=x\sqrt{(x-2)^2+(y-0)^2}-1=|x|(x2)2+(y0)2=x+1\sqrt{(x-2)^2+(y-0)^2}=|x|+1x24x+4+y2=x2+2x+1x^2-4x+4+y^2=x^2+2|x|+1y2=4x+2x3y^2=4x+2|x|-3

Since y20,yR,y^2\geq0, y\in \R, we take x=x,x0|x|=x, x\geq0

y2=4x+2x3y^2=4x+2x-3y2=6x3y^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)(2,0) is the equaion of the parabola


y2=6x3y^2=6x-3

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