Questions on Geometry: Length, distance, coordinates, metric length answered by real tutors!

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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.: 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.
Answer by Alan3354(1926) About Me  (Show Source):
You can put this solution on YOUR website!
It's a parabola:
y = -(7x^2 - 28x + 208)/72
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.: 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.
Answer by Edwin McCravy(2190) About Me  (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.

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