Thread: analytic geometry

The center and radius of circumference M are (0,0) and 10, respectively. The center and radius of circumference P are (8,0) 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.

THANK YOU IN ADVANCE!!!!!

Let circumference Q be of radius r. Its center, Q(x,y)

Circumference M has a radius of 10, and its center is M(0,0)
Circumference P has a radius of 2, and its center is P(8,0)

MQ = sqrt[(x-0)^2 +(y-0)^2] = 10 -r
So, r = 10 - sqrt[x^2 +y^2] --------------------(i)

PQ = sqrt[(x-8)^2 +(y-0)^2] = 2 +r
So, r = sqrt[x^2 -16x +64 +y^2] -2 --------------(ii)

r = r,
10 - sqrt[x^2 +y^2] = sqrt[x^2 -16x +64 +y^2] -2
12 - sqrt[x^2 +y^2] = sqrt[x^2 -16x +64 +y^2]
Square both sides,
144 -24sqrt[x^2 +y^2] +x^2 +y^2 = x^2 -16x +64 +y^2
-24sqrt[x^2 +y^2] = -16x -80
Divide both sides by -8,
3sqrt[x^2 +y^2] = 2x +10
Square both sides,
9[x^2 +y^2] = 4x^2 +40x +100
9x^2 +9y^2 = 4x^2 +40x +100
5x^2 -40x +9y^2 = 100
5[x^2 -8x] +9y^2 = 100
5[x^2 -8x +16 -16] +9y^2 = 100
5[x^2 -8x +16] +9y^2 = 100 +80
5[(x -4)^2] +9y^2 = 180 ---------------------an ellipse.
Divide both sides by 180,
[(x -4)^2]/36 +(y^2)/20 = 1
[(x -4)^2]/(6^2) +[(k -0)^2]/(2sqrt(5))^2 = 1 -----------***

That is the the equation of the locus of the centers of the circles of the Q kind.
It is an ellipse whose center is at (4,0), whose semi-major axis is 6, and whose semi-minor axis is 2sqrt(5).