SOLUTION: A moving circle is tangent to the x-axis and to a circle of radius one with center at (2,-6). Find the equation of the locus of the center of the moving circle.

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Question 169559: A moving circle is tangent to the x-axis and to a circle of radius one with center at (2,-6). Find the equation of the locus of the center of the moving circle.
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
It's a parabola:


Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Warning: Alan's equation
is incorrect!
Edwin's solution:

A moving circle is tangent to the x-axis and to a circle of radius one with center at (2,-6). Find the equation of the locus of the center of the moving circle.

Let's draw the given circle with center (2,-6) and radius 1.

 

Next let's draw in any arbitrary circle which is tangent 
to both that circle and the x-axis and label its center 
as the general point (x,y) :

 

Draw a line segment connecting the center of this circle
(x,y), to the center of the given circle.



Use the distance formula to find an expression
for the distance between them:





That distance is the radius of the arbitrary circle
plus 1.

Now let's draw a radius of the arbitrary circle 
vertically up to the x-axis.



That vertical radius of the arbitrary circle must 
equal to -y in length because y must be negative, 
and so -y must be positive.

Distance between centers = 

          radius of arbitrary circle +
 
                      radius of the given circle.

So 



Squaring both sides:











That is the equation you are looking for.

Now let's draw the graph of that.

It is obviously a parabola opening
downward:



Notice that any point we pick on that parabola,
draw a circle with that center tangent to the x-axis, 
it will automatically be tangent to the given circle:



Edwin

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