SOLUTION: find the equation satisfied by the centers of the set of circles such that each is tangent to x^2+y^2=9 and to x=5. thank you P.s Is my answer y^2=-4(x-1) and y^2=-16(x-4) correct

Algebra ->  Length-and-distance -> SOLUTION: find the equation satisfied by the centers of the set of circles such that each is tangent to x^2+y^2=9 and to x=5. thank you P.s Is my answer y^2=-4(x-1) and y^2=-16(x-4) correct      Log On


   

Question 1131587: find the equation satisfied by the centers of the set of circles such that each is tangent to x^2+y^2=9 and to x=5. thank you
P.s Is my answer y^2=-4(x-1) and y^2=-16(x-4) correct?
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2=9 and to x=5. thank you
P.s Is my answer y^2=-4(x-1) and y^2=-16(x-4) 
 

The larger red circle is x²+y²=9.  The smaller black circle is a typical 
circle tangent to the large circle and to the vertical line x=5. Its center
is a point (x,y) on the desired equation. 

[blue radius] = 3

[green radius] = 5-x = [red radius]

[blue radius] + [red radius] = [distance from (0,0) to (x,y)] = √x²+y²)

So the un-simplified equation is:

3 + (5-x) = √x²+y²  

8 - x = √x²+y²

Squaring both sides:

(8 - x)² = (√x²+y²)²

64 - 16x + x² = x² + y²

64 - 16x = y²

y² = -16x + 64

y² = -16(x - 4),  a parabola

You got that for one of your answers.

[Where did you get y² = -4(x-1)?]

Now we put that parabola on the other one graph, and draw a couple more
circles, using a different scale:



Edwin