Question #250674

Find the equation of the locus of the center of a moving circle tangent to the y-axis and to a circle with a radius 2 with center at (8, 2).


1
Expert's answer
2021-10-18T15:56:03-0400

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 2 with center at (8, 2).

The distance from point PP to the y-axis is x|x|. The distance from the point PP to a center (8,2)(8, 2) is

(x8)2+(y2)2.\sqrt{(x-8)^2+(y-2)^2}. The distance from the point PP to a circle is (x8)2+(y2)22.\sqrt{(x-8)^2+(y-2)^2}-2.

Then

(x8)2+(y2)22=x\sqrt{(x-8)^2+(y-2)^2}-2=|x|


(x8)2+(y2)2=x+2\sqrt{(x-8)^2+(y-2)^2}=|x|+2

x216x+64+(y2)2=x2+4x+4x^2-16x+64+(y-2)^2=x^2+4|x|+4

(y2)2=16x+4x60(y-2)^2=16x+4|x|-60

Since (y2)20,yR,(y-2)^2\geq0, y\in \R, we take x=x,x0|x|=x, x\geq0


(y2)2=16x+4x60(y-2)^2=16x+4x-60

(y2)2=20x60(y-2)^2=20x-60

The equation of the locus of the center of a moving circle tangent to the y-axis and to a circle with a radius 2 with center at (8, 2) is the equaion of the parabola


(y2)2=20x60(y-2)^2=20x-60


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