Four circles are arranged as shown in the figure below. They're numbered from $1$ to $4$.
If the diameter of circle $C_2$ is equal to $9$, and $ PT = 6 , QT = 3 \sqrt{5} $. It is also given that $UV = 18$. Find the radius of the red circle.
My attempt:
Let the center $C_2 = (0,0)$, and let $UP = a$ and $QV = b$, and let $r_i$ be the radius of the $i$-th circle, then we have $r_1, r_3, r_4$ unknown , while $r_2 = 4.5 $. The coordinates of $C_i$ are
$ C_1 = (-4.5 – a, r_1) $
$C_2 = ( 0, 0) $
$ C_3 = (4.5 + b , r_3 )$
$ C_4 = (x_4, y_4) $
Now we write $5$ equations stemming from the distance formula relating the squared distances between pairs of centers chosen out the $4$ circles, and the square of the sum of their radii.
Taking the following pairs of circles: $(1, 2), (2, 3), (1, 4), (2, 4), (3, 4) $ gives us
$ (4.5 + a)^2 + r_1^2 = (r_1 + 4.5)^2 \tag{1} $
$ (4.5 + b)^2 + r_3^2 = (r_3 + 4.5)^2 \tag{2} $
$ (4.5 + a + x_4)^2 + (y_4 – r_1)^2 = (r_1 + r_4)^2 \tag{3}$
$ x_4^2 + y_4^2 = (4.5 + r_4)^2 \tag{4}$
$ (4.5 + b – x_4)^2 + (y_4 – r_3)^2 = (r_3 + r_4)^2 \tag{5}$
In addition to these five equations, we know that point $T$ is given by
$ T = P + PT ( \cos \phi, \sin \phi ) $
Where $ \cos \phi = \dfrac{PT}{PQ} = \dfrac{6}{9} = \dfrac{2}{3} $, therefore, $\sin \phi = \dfrac{\sqrt{5}}{3} $
Therefore,
$ T = (-4.5, 0) + 6 ( \dfrac{2}{3} , \dfrac{\sqrt{5}}{3} ) = (-0.5 , 2 \sqrt{5} ) $
It follows that
$ C_4 = \left(\dfrac{ r_4 + 4.5 }{4.5}\right) T $
And this reads
$ x_4 = – 0.5 (1 + \dfrac{2}{9} r_4 ) \tag{6}$
$ y_4 = 2 \sqrt{5} (1 + \dfrac{2}{9} r_4 ) \tag{7}$
With this equation $(4)$ becomes redundant.
And finally we have
$ a + b = 9 \tag{8}$
Now, equations $(1),(2), (3), (5)$ after expansion, become:
$9 a+ a^2 = 9 r_1 \tag{9} $
$ 9 b + b^2 = 9 r_3 \tag{10} $
$ 4.5^2 + a^2 + x_4^2 + 9 a + 9 x_4 + 2 a x_4 + y_4^2 – 2 y_4 r_1 = 2 r_1 r_4 + r_4^2 \tag{11} $
$ 4.5^2 + b^2 + x_4^2 + 9 b – 9 x_4 – 2 b x_4 + y_4^2 – 2 y_4 r_3 = 2 r_3 r_4 + r_4^2 \tag{12} $
Substituting equations $(9)$ and $(10)$ into $(11)$ and $(12)$ gives
$ 4.5^2 + 9 r_1 + x_4^2 + y_4^2 + 9 x_4 + 2 a x_4 – 2 y_4 r_1 = 2 r_1 r_4 + r_4^2 \tag{13}$
$ 4.5^2 + 9 r_3 + x_4^2 + y_4^2 – 9 x_4 – 2 b x_4 – 2 y_4 r_3 = 2 r_3 r_4 + r_4 ^ 2 \tag{14}$
So our system of equations is given by equations $(6), (7), (8)$ which are linear and equations $(9), (10), (13), (14)$ which are quadratic.
What remains is to how to solve these equations.
So the question here is: Are there further simplifications that can render this system of equations solvable?
I appreciate your input on this.
Edit: I found a reduction that can prove helpful. Subtracting equation $(10)$ from $(9)$, we get
$ 9 (a – b) + (a – b)(a+b) = 9 r_1 – 9 r_3 $
But $ a + b = 9 $, therefore,
$ 18 (a – b) = 9 (r_1 – r_3) $
i.e.
$ 2 (a – b) = r_1 – r_3 \tag{15} $
which is linear in the variables. So now we have $4$ linear equations which are equations $(6),(7),(8),(15)$ and three quadratic equations which are equations $(9), (13), (14)$.
Best Answer
Let's call the radii, respectively, $r_1, r_2, r_3, r_4$, where $r_2=4.5$ is already known to us. Moreover, assume $\angle C_4C_2P= \theta$ and $|UP|=x.$ By the length values given, it's very easy (please be aware that $C_4$, $C_2$, and $T$ are collinear) to get $\cos \theta=\frac{1}{9}$ and $\sin \theta =\frac{4\sqrt 5}{9}$.
Now, note that:
$$|C_1C_4|=r_1+r_4, \\ |C_2C_4|=4.5+r_4, \\ |C_4M|=(4.5+r_4)\sin \theta=(4.5+r_4)\frac{4\sqrt 5}{9}.$$
And,
$$=(|C_4M|-r_1)^2+(|UC_2|-|C_2C_4| \cos \theta)^2=|C_1C_4|^2.$$
Hence,
$$((4.5+r_4)\frac{4\sqrt 5}{9}-r_1)^2+(4+x-\frac{r_4}{9})=(r_1+r_4)^2. \ \ \ \ (1)$$
Similarly,
$$((4.5+r_4)\frac{4\sqrt 5}{9}-r_3)^2+(14-x+\frac{r_4}{9})=(r_3+r_4)^2. \ \ \ \ (2)$$
On the other hand, note that:
$$|C_1C_2|^2-|C_1U|^2=|C_2U|^2.$$
Hence,
$$(4.5+r_1)^2-r_1^2=(4.5+x)^2. \ \ \ \ (3)$$ Similarly,
$$(4.5+r_3)^2-r_3^2=(13.5-x)^2. \ \ \ \ (4)$$
By some help from Wolfram Alpha to solve $(1), (2), (3), $ and $(4)$, we have:
$$r_4 \approx 2.31, \\ r_3 \approx 8.29, \\r_1 \approx 5.32.$$
Remark 1: In the picture above we indeed have: $r_1=a, r_3=c,$ and $r_4=d$.
Remark 2: The numbers are very compatible with the figure posted by Jean Marie.