HEEELLLPPP! calculus!?

Question

Please help! i fill like my brain is going to come out of my brain! i dont know how to solve this i keep trying but i cant!
The center and radius of circumference M are (0,0) and 10, respectively. The center and radius of circumference P are (8,2) and 2, respectively. Circumference Q is internally tangent to circumference M and externally tangent to circumference P.Of course that there are infinitely many circumferences Q which have this property. Determine and graph the equation of the curve that contains the centers of all the circumferences of the Q kind.

Answer 1

the distance between centers of 2 tangent circumferences =
sum of radius or difference (tangent exterbnally or internally)

If (x,y,)is the centers of such a cf.
(x-8)^2+(y-2)^2= (r+2)^2
x^2+y^2 -16x-4y +64 = r^2+4r (a)

x^2+y^2 =(10-r)^2
x^2+y^2= r^2 -20r+100 (b)
16x+4y-64 = -24r+100 so r = -(16x+4y-164)/24 =
=(-4x-y+41)/6
so x^2+y^2 -16x-4y+64 = (-4x-y+41)^2/36 +2/3(-4x-y+41)
This is the equation you are looking for
You can do the operations

Answer 2

Alright, I'll take a shot at this.

You've got a point (x,y) that is equidistant from the closest points of M and P. It is inside M but outside P. This point is a solution, right? A potential center of Q?

The closest point of P to (x,y) will lie directly between (x,y) and (8,2). I'm going to rename (x,y) point A, if you don't mind. So, the distance between A and P is sqrt((x - 8)^2 + (y - 2)^2) - 2.

The closet point of M to A will be on the ray starting at the origin and going through A. The distance from this point to A will be the radius (10) minus the distance of A from (0,0). So, this distance will be 10 - sqrt(x^2 + y^2).

And if A must be equidistant from M and P, then the two distances I've just found must be equivalent. Therefore, the equation you seek is

sqrt((x - 8)^2 + (y - 2)^2) - 2 = 10 - sqrt(x^2 + y^2)

Or

12 = sqrt((x - 8)^2 + (y - 2)^2) + sqrt(x^2 + y^2)

Graphing this, naturally, is a beast, since it is not a function. And there's no way I can really show the graph to you in type. All I can tell you, and you probably guessed this already, is that it forms a horseshoe opening to the east, with the ends at the intersections of M and P.

EDIT - I believe the "horseshoe" in question to be an ellipse, just with one part missing on the west end. The kicker is that the ellipse is not on axes parallel to the x and y axes, but rotated just a few degrees counterclockwise.