Question

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?

Answer 1

`.5`

Explanation:

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`