Question 169559
<font size = 7 color = "red"><b>Warning: Alan's equation
is incorrect!
Edwin's solution:</b></font>

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.
<pre><font size = 4 color = "indigo"><b>
Let's draw the given circle with center (2,-6) and radius 1.

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10), locate(2,-6,"(2,-6)"),
circle(2,-6,1), line(1.9,-6,2.1,-6),line(2,-5.9,2,-6.1) 
)}}} 

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) :

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10), locate(2,-6,"(2,-6)"),
circle(2,-6,1), line(1.9,-6,2.1,-6),line(2,-5.9,2,-6.1), 
circle(6,-3.642857143,3.642867143), 
line(6,-3.642857143-.1,6,-3.642867143+.1),
line(6.1,-3.642857143,5.9,-3.642867143),
locate(6,-3.65,"(x,y)")
)}}} 

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

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10), locate(2,-6,"(2,-6)"),
circle(2,-6,1), line(1.9,-6,2.1,-6),line(2,-5.9,2,-6.1), 
circle(6,-3.642857143,3.642867143), 
line(6,-3.642857143-.1,6,-3.642867143+.1),
line(6.1,-3.642857143,5.9,-3.642867143),
locate(6,-3.65,"(x,y)"), line(6,-3.642857143,2,-6)  
)}}}

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

{{{d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}
{{{d=sqrt((x-2)^2+(y-(-6))^2)}}}
{{{d=sqrt((x-2)^2+(y+6)^2)}}}

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.

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10), locate(2,-6,"(2,-6)"),
circle(2,-6,1), line(1.9,-6,2.1,-6),line(2,-5.9,2,-6.1), 
circle(6,-3.642857143,3.642867143), 
line(6,-3.642857143-.1,6,-3.642867143+.1),
line(6.1,-3.642857143,5.9,-3.642867143),
locate(6,-3.65,"(x,y)"), line(6,-3.642857143,2,-6),  
line(6,-3.642867143,6,0)
)}}}

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 

{{{sqrt((x-2)^2+(y+6)^2) = -y+1}}}

Squaring both sides:

{{{(x-2)^2+(y+6)^2 = (-y+1)^2}}}

{{{x^2-4x+4+y^2+12y+36=y^2-2y+1}}}

{{{x^2-4x+14y+39=0}}}

{{{14y=-x^2+4x-39}}}

{{{ y=(-x^2+4x-39)/14 }}}

That is the equation you are looking for.

Now let's draw the graph of that.

It is obviously a parabola opening
downward:

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10,(4x-x^2-39)/14), locate(2,-6,"(2,-6)"),
circle(2,-6,1), line(1.9,-6,2.1,-6),line(2,-5.9,2,-6.1), 
circle(6,-3.642857143,3.642867143), 
line(6,-3.642857143-.1,6,-3.642867143+.1),
line(6.1,-3.642857143,5.9,-3.642867143),
locate(6,-3.65,"(x,y)"), line(6,-3.642857143,2,-6),  
line(6,-3.642867143,6,0)
)}}}

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:

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10,(4x-x^2-39)/14), locate(2,-6,"(2,-6)"),
circle(2,-6,1), line(1.9,-6,2.1,-6),line(2,-5.9,2,-6.1), 
circle(6,-3.642857143,3.642867143), 
line(6,-3.642857143-.1,6,-3.642867143+.1),
line(6.1,-3.642857143,5.9,-3.642867143),
locate(6,-3.65,"(x,y)"), line(6,-3.642857143,2,-6),  
line(6,-3.642867143,6,0),circle(-5,-6,6), line(-4.9,-6,-5.1,-6),
line(-5,-6.1,-5,-5.9)
)}}}

Edwin</pre>