Question

can anyone help me solve this hard problem please?

Answer 1

Relevant:
http://mathworld.wolfram.com/TangentCircles.html

Answer 2

Introducing some coordinate axis.

Answer 3

(x-x1)^2 + (y-y1)^2 - (r+-r1)^2 = 0
(x-x2)^2 + (y-y2)^2 - (r+-r2)^2 = 0
(x-x3)^2 + (y-y3)^2 - (r+-r3)^2 = 0

Should give 3 equations in 3 unknowns...

Answer 4

hmm then?

Answer 5

Circle 1 is the radius 1 circle which is centered at (0,1)
Circle 2 is the radius 2 circle which is centered at (0, -2)
Circle 3 is the radius 3 circle which is centered at (0, -1)
Also, from this quote "If the center of the second circle is outside the first, then the sign corresponds to externally tangent circles and the sign to internally tangent circles." I'm choosing to rewrite (r+-r1) as r-r1 and r-r2, since we only care about the externally tangent circles for the first two circles. Similarly, for the 3rd circle, I choose to use just the + sign, since we want those to circles to be internally tangent. That gives:

(x-0)^2 + (y-1)^2 - (r-1)^2 = 0
(x-0)^2 + (y-(-2))^2 - (r-2)^2 = 0
(x-0)^2 + (y-(-1))^2 - (r+3)^2 = 0

Answer 6

Simplifies to:
x^2 +y^2-2y -r^2 +2r=0
x^2 +y^2 +4y -r^2 +4r =0
x^2 +y^2 +2y -r^2 -6r -8 =0

Answer 7

Hope I didn't make any mistakes on that algebra...
How easily solved is that system? We only need to solve it for r...

Answer 8

Hey, I'm almost there. Notice, all three of those equations involve x^2 + y^2 -r^2...
So I can subtract the equations from one another to cancel those terms! =D

Answer 9

(1) x^2 +y^2-2y -r^2 +2r=0
(2) x^2 +y^2 +4y -r^2 +4r =0
(3) x^2 +y^2 +2y -r^2 -6r -8 =0
(1)-(2) gives -6y -2r = 0
(1)-(3) gives -4y +8r = 0
SOLVE FOR R. OH BABY.

Answer 10

the r gives a negative number, is that normal?

Answer 11

Does it? haha I actually got r = 0...

Answer 12

And no... r shouldn't be negative. Lol.
I might be on the wrong track here. Or I may have made one stupid mistake somewhere.

Answer 13

it gives me r=-6/7

Answer 14

Do you have work for that? I'm not seeing it.

Answer 15

Ooooh. I see my mistake.

Answer 16

we love you smoothmath

Answer 17

Just a minute.

Answer 18

(1) x^2 +y^2-2y -r^2 +2r=0
(2) x^2 +y^2 +4y -r^2 +4r =0
(3) x^2 +y^2 +2y -r^2 -6r -8 =0
(1)-(2) gives
-6y -2r = 0
(1)-(3) gives
-4y -8r +8 = 0 (This is where I made my last mistake)

Solving those two equations for r...
y = (-2/6)r
-4((-2/6)r) - 8r = -8
(8/6r) - 8r = -8
(8/6 - 8) r = -8
(-40/6)r = -8
r = 48/40

Answer 19

Hells yes.

Answer 20

I love you too mathaddict.

Answer 21

Still gotta write 48/40 is reduced form and then do numerator + denominator for some arbitrary reason LOLS.
I trust you can do that? haha

Answer 22

Math is such a thrill.

Answer 23

thanks smoothmath

Answer 24

@SmoothMath , i think (1)-(3) gives -4y +8r +8 = 0 rather than -4y -8r +8 = 0
this way, r = -6/7, which is a negative value, is it normal?

Answer 25

Hmph. Yeah I'm getting the same thing. I don't know if it's normal or not. I suppose a positive radius would be the same as a negative radius... =/

Answer 26

Lemme go back and check my process though...

Answer 27

(x-0)^2 + (y-1)^2 - (r-1)^2 = 0
(x-0)^2 + (y+2)^2 - (r-2)^2 = 0
(x-0)^2 + (y+1)^2 - (r+3)^2 = 0

x^2 +(y^2 -2y +1) - (r^2 -2r +1) = 0
x^2 +(y^2 +4y +4) - (r^2 -4r +4) = 0
x^2 +(y^2 +2y +1) - (r^2 +6r +9) = 0

x^2 +y^2 -r^2 -2y +2r =0
x^2 +y^2 -r^2 +4y +4r =0
x^2 +y^2 -r^2 +2y -6r -8 =0

Same result.
Blah.

Answer 28

doesn't matter, really appreciated for your time, i'll look at the rest by myself ;)

Answer 29

Here are two ways to solve this. Personally, I like smoothway's approach the best.

Answer 30

Nice =D