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Three semi-circles have their centers as follows: $(0,0), (2, 0), (-4, 0)$ with radii of $6, 4, 2$, respectively. Determine the radius and center coordinates of a circle that is tangent to all three semi-circles.

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My attempt:

Let $C_1 = (x_1, y_1) = (0,0), C_2 = (x_2, y_2)= (2, 0), C_3 = (x_3,y_3) = (-4, 0) $ and let $r_1 = 6 , r_2 = 4, r_3 = 2 $.

Further, let $C_4 = (x_4, y_4)$ be the center of the required circle, and $r_4$ be its radius, then by connecting the centers of the fourth circle with each of the first three, we can write the following three quadratic equations:

$ (x_4 - x_1)^2 + (y_4 - y_1)^2 = (r_1 - r_4)^2 $

$ (x_4 - x_2)^2 + (y_4 - y_2)^2 = (r_2 + r_4) ^ 2$

$ (x_4 - x_3) ^ 2 + (y_4 - y_3) ^ 2 = (r_3 + r_4) ^ 2 $

My question is, how to solve these three equations for the unknowns $x_4, y_4, r_4$?

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    $\begingroup$ Hint: the radii of the four circles satisfy Descrate's theorem: $$\left(-\frac1{r_1} + \frac1{r_2} + \frac1{r_3} + \frac1{r_4}\right)^2 = 2\left(\frac1{r_1^2} + \frac1{r_2^2} + \frac1{r_3^2} + \frac1{r_4^2}\right)$$ $\endgroup$ Commented Nov 6, 2023 at 9:27
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    $\begingroup$ @JuvHuffpuff $(r_1-r_4)^2$ is correct because circle 4 is tangent to inside of circle 1. $\endgroup$ Commented Nov 6, 2023 at 9:29
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    $\begingroup$ The formula given by @achille hui can be complemented by another one giving (in a complex numbers setting) the centers of the circles (see formula (2) here) $\endgroup$
    – Jean Marie
    Commented Nov 6, 2023 at 9:40
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    $\begingroup$ Subtracting the first equation from the other two, you get two linear equations which you can use to express $x_4$ and $y_4$ as a function of $r_4$. Plug then those expressions into the first equation. $\endgroup$ Commented Nov 6, 2023 at 10:32
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    $\begingroup$ That's okay. I am not gonna solve it by hand. I'll using my math library to solve the linear system and the quadratic system that follows. $\endgroup$ Commented Nov 6, 2023 at 13:00

1 Answer 1

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As @intelligentipauca suggested, I started by subtracting the second equation from the first, this gives

$x_4 (-2 x_1 + 2 x_2 ) + y_4 (- 2 y_1 + 2 y_2 ) + x_1^2 - x_2^2 + y_1^2 - y_2^2 = r_4 (- 2 r_1 - 2 r_2 ) + r_1^2 - r_2^2 $

And similarly, subtracting the third equation from the first equation gives

$x_4 (-2 x_1 + 2 x_3 ) + y_4 (- 2 y_1 + 2 y_3 ) + x_1^2 - x_3^2 + y_1^2 - y_3^2 = r_4 (- 2 r_1 - 2 r_3 ) + r_1^2 - r_3^2 $

Now we have a linear system of two equations in $3$ unknowns.

The system can be expressed as follows

$ A X = B $

where $X = [x_4, y_4, r_4]^T $

and

$A = \begin{bmatrix}-2 x_1 + 2 x_2 && - 2 y_1 + 2 y_2 && 2 r_1 + 2 r_2 \\ -2 x_1 + 2 x_3 && - 2 y_1 + 2 y_3 &&2 r_1+ 2 r_3\end{bmatrix} $

and

$B = \begin{bmatrix} -x_1^2 + x_2^2 - y_1^2 + y_2^2 + r_1^2 - r_2^2 \\ -x_1^2 + x_3^2 - y_1^2 + y_3^2 + r_1^2 - r_3^2 \end{bmatrix}$

Plugging the given values of $x_1, x_2, x_3, y_1, y_2, y_3, r_1, r_2, r_3 $, gives us

$A = \begin{bmatrix} 4 && 0 && 20 \\ -8 && 0 && 16 \end{bmatrix} $

$ B = \begin{bmatrix} 24 \\ 48 \end{bmatrix} $

The solution of which is

$ X = \dfrac{1}{7} \begin{bmatrix} -18 \\ 0 \\ 12 \end{bmatrix} + \lambda \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $

So that,

$ C_4 = \dfrac{1}{7} \begin{bmatrix} -18 \\ 0 \end{bmatrix} + \lambda \begin{bmatrix} 0 \\ 1 \end{bmatrix} $

And

$ r_4 = \dfrac{12}{7} $

Pluggin these into

$ (C_4 - C_1)^T (C_4 - C_1) = (r_1 - r_4)^2 $

Gives

$ \lambda = \dfrac{24}{7} $

So that

$ C_4 = \dfrac{1}{7} \begin{bmatrix} -18 \\ 24 \end{bmatrix} $

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    $\begingroup$ In this particular case the coefficient of $y_4$ vanishes in both linear equations. Hence you have a linear system with only two unknowns ($r_4$ and $x_4$) which can be solved. No need to use Descartes' formula. $\endgroup$ Commented Nov 6, 2023 at 22:08
  • $\begingroup$ Thanks @Intelligentipauca, I've modified my solution to avoid the calculation of $r_4$ based of Descartes' formula. $\endgroup$ Commented Nov 9, 2023 at 12:07

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