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Question 1146375: Given a circle of radius of 4 and center at (6, 8). Find the locus of the center of all circles that is tangent to the x - axis and the given circle.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The required conditions for each point in the locus are
(1) The radius of the circle is the y value; and
(2) The radius of the circle plus 4 is the distance between the point and the center of the circle.
Those two requirements give us, for any point (x,y) on the locus,
Here is another path to the same solution.
With the circle with center (6,8) and radius 4, the point on the circle closest to the x-axis is (6,4); then clearly one point on the locus is (6,2) -- halfway between (6,4) and the x-axis.
It should also by clear by symmetry that the locus will be symmetrical about the line x=6; any point on the locus "a" units to the right of the line x=6 will be mirrored by a point on the locus "a" units to the left of the line x=6.
So the locus is symmetrical about the line x=6; it is therefore a parabola with vertex (6,2) and an equation of the form
To determine the value of the constant a, note that a point on the locus to the right of x=6 will have a y value of 8 and an x value of 6 (center of circle) plus 4 (radius of circle) plus 8 (equal to the y value).
That gives us the point (18,8). Plugging those values in the equation of the parabola gives us the value of a:
And again we find the equation of the locus i
or
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