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Question 152311: 28.) A moving circle is tangent to the x-axis and to a circle of radius 1 with center at (2,-6). Find the equation of the locus of the center of the moving circle.
Ans. 
This question came with the answer included above. The teachers at my school only taught us how to solve for the equation of the locus of moving points not moving circles. Not even my classmates can help me with it and I have to solve it...
Answer by oscargut(2103)   (Show Source):
You can put this solution on YOUR website! Let x and y the center of the moving circle
then distance between (x,y)and x-axis = distance between (x,y)and (2,-6)-1
Obs: y<0
d((x,y),(x,0))=d((x,y),(2,-6))-1
-y+1=d((x,y),(2,-6))
[-y+1]^2=[d((x,y),(2,-6))]^2
y^2-2y+1=(x-2)^2+(y+6)^2
y^2-2y+1=x^2-4x+4+y^2+12y+36
-2y+1=x^2-4x+4+12y+36
-14y-x^2+4x-39=0
(check your solution please)
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