The lengths of the tangents from any point on the circle #15x^2 + 15y^2 -48x +64 y =0# to the two circles #5x^2 + 5y^2 -24x +32y +75 =0# and #5x^2 + 5y^2 -48x +64y +300 =0# are in the ratio?

To give an idea of what is intended, take a look at the radical axis, the locus of points where the tangents of two circles meet at the ratio 1:1

The Radical Axis

First let's rewrite the 3 circles' equations
Showing the way with the first equation:

#15x^2+15y^2-48x+64y=0#
#x^2+y^2-16/5x+64/15y=0#
#-> -2x_0=-16/5# => #x_0=8/5#
#-> -2y_0=64/15# => #y_0=-32/15#
#-> x_0^2+y_0^2-r^2=0# => #r^2=64/25+1024/225# => #r=8/3#
#circ_0 -> (x-8/5)^2+(y+32/15)^2=(8/3)^2#

Following this way

#circ_1 -> (x-12/5)^2+(y+16/5)^2=1^2#
#circ_2 -> (x-24/5)^2+(y+32/5)^2=2^2#

Graphing these three circles we'll get something like the figure below

Please remark that the circles' centers (#C_0, C_1 and C_2#) are colinear
The slope of the line in which they are is
#C_0C_1 -> k=(-16/5+32/15)/(12/5-8/5)=(-16/15)/(4/5)=-4/3# => #y=-(4/3)x#
#tan phi=-4/3# => #phi=-53.13^@#

Finding the point P (combining equation of circle-0 and equation of the line, #y=-(4/3)x#)

#(x-8/5)^2+(y+32/15)^2=(8/3)^2#
#(x-8/5)^2+(-4/3x+32/15)^2=64/9#
#x^2-16/5x+64/25+16/9x^2-256/45x+1024/225-64/9=0#
#25/cancel(9)x^2-80/cancel(9)x=0# => #x_1=0# (not valid); #x_2=16/5#
#-> y=-(4/3)x=-(4/3)16/5# => #y=-64/15#
#-> P(16/5,-64/15)#

Next step will be to translate point #P# to origin #O (0,0)#
With this translation the circles' centers

#C_0 (8/5,-32/15), C_1 (12/5, -16/5) and C_2 (24/5, -32/5)#
become
#C_0 (-8/5, 32/15), C_1 (-4/5, 16/15) and C_2 (8/5, -32/15)#

Now these points will rotate #-phi=53.13^@# about the origin towards the x-axis.
Since the line common to the three circles' centers will become coincident to the new x-axis (#y=0#), there's no need to calculate y' (it will be equal to 0).

#x'=x*cos(-phi)-y*sin(-phi)#
#x'_0=(-8/5)*cos53.15^@-(32/15)*sin53.15^@=-2.667=-8/3#
In this way
#x'_1=-1.333=-4/3#
#x'_2=2.667=8/3#

So, after the translation and rotation

#C_0 (-8/3,0), C_1 (-4/3,0) and C_2 (8/3,0)#

(see figure 2, below)

Consider special case of Figure 3

In the special case we can see that

#AD=C_1A-r_1=4/3-1=1/3#
#AB^2=AD*(AD+2r_1)=(1/3)(1/3+2*1)=7/9# => #AB=sqrt(7)/3#
#AH=C_2A-r_2=8/3-2=2/3#
#AG^2=AH*(AH+2r_2)=(2/3)(2/3+2*2)=28/3# => #AG=2sqrt(7)/3#
So the ratio #(AB)/(AG)=1/2#

Now for the general proof
Consider Figure 4

As we saw above

#AB^2=AD*(AD+2r_1)#
Or
#AB^2=[(AD+r_1)-r_1]xx[(AD+r_1)+r_1]#
And
#AG^2=[(AH+r_2)-r_2]xx[(AH+r_2)+r_2]#

The equation of the circle-0 in polar coordinates is

#r=-2a*cos theta# => #r=-16/3*cos theta#

Points #C_1 and C_2# in polar coordinates are

#C_1 (4/3, pi) and C_2 (8/3,0)#

The distance between two points in polar coordinates is

#d=sqrt (r_a^2+r_b^2-2*r_a*rb*cos(theta_a-theta_b))#

So the distance between #C_1# and a point A in the circle-0 (distance #AD+r_1#) is
#d_(10)=sqrt(256/9*cos^2 theta+16/9+2*16/3*cos theta*4/3*cos(theta-pi)#
#d_(10)=sqrt(256cos^2 theta+16-128cos^2 theta)/3#
#d_(10)=4/3sqrt(8cos^2 theta+1)#

#AB^2=[4/3sqrt(8cos^2 theta +1)-1][4/3sqrt(8cos^2 theta+1)+1]#
#AB^2=[16/9(8cos^2 theta +1)-1]=(128cos^2 theta +7)/9#

And the distance between #C_2# and the same point A in the circle-0 (distance #AH+r_2#) is
#d_(20)=sqrt(256/9cos^2 theta+64/9+2*16/3*cos theta *8/3*cos(theta-0))#
#d_(20)=sqrt(256cos^2 theta+64+256cos^2 theta)/3#
#d_(20)=8/3sqrt(8cos^2 theta+1)#

#AG^2=[8/3sqrt(8cos^2 theta+1)-2][8/3sqrt(8cos^2 theta +1)+2]#
#AG^2=[64/9(8cos^2 theta +1)-4]=(512cos^2 theta +28)/9#
#AG^2= 4*(128cos^2 theta +7)/9#

So
#(AB)^2/(AG)^2=(cancel((128cos^2 theta +7)/9))/(4*cancel((128cos^2 theta+7)/9))=1/4#

And the ratio is
#(AB)/(AG)=1/2#