Circle touching three tangential circles

Question

The circles $C_1,C_2$ and $C_3$ with radii $1,2$ and $3$, respectively, touch each other externally. The centres of $C_1$ and $C_2$ lie on the $x$-axis, while $C_3$ touches them from the top. Find the ordinate of the centre of the circle that lies in the region enclosed by the circles $C_1,C_2$ and $C_3$ and touches all of them.

I had been trying to do the sum by assuming the centre of the first circle at $(0,1)$ but it did not help.

Answer 1

You may assume that the center of $C_1$ is at $(0,0)$ and the center of $C_3$ is at $(0,4)$.

Now, the unknown point is at, say $(x,y)$. Find it from the equations

$$x^2 + y^2 = (1 + r)^2$$ $$ x^2 + (4-y)^2 = (3 + r)^2$$ $$ (3-x)^2 + y^2 = (2 + r)^2$$

where $r$ is the radius of the inscribed circle. You have 3 equations with 3 unknowns, should be able to finish it from here.

Corrected: I've had a brain sneeze on the RHS.

Answer 2

here is my nice answer:

let $C$ be the center of enclosed circle then its radius $r$ is given by set formula by HC Rajpoot $$r=\frac{abc}{2\sqrt{abc(a+b+c)}+ab+bc+ca}$$ where, $a=1, b=2, c=3$ are radii of three externally touching circles then $$r=\frac{1\times2\times 3}{2\sqrt{1\times 2\times 3(1+2+3)}+1\times 2+2\times 3+3\times 1}=\frac{6}{23}$$ now, drop a perpendicular $CN$ of length $y$ from center $C$ to the x-axis to get a right triangle $C_1NC$ in which hypotenuse $C_1C=1+\frac6{23}=\frac{29}{23}$. apply pythagorean $$C_1N=\sqrt{(C_1C)^2-(CN)^2}=\sqrt{\left(\frac{29}{23}\right)^2-y^2}\ \ \ \ ..........(1)$$ similarly, in right triangle $CNC_2$ in which hypotenuse
$C_2C=2+\frac6{23}=\frac{52}{23}$. apply pythagorean $$NC_2=\sqrt{(C_2C)^2-(CN)^2}=\sqrt{\left(\frac{52}{23}\right)^2-y^2}\ \ \ \ ........(2)$$ since, circle $C_1$ & $C_2$ are touching hence, $$C_1N+NC_2=C_1C_2=1+2=3$$ $$\sqrt{\left(\frac{29}{23}\right)^2-y^2}+\sqrt{\left(\frac{52}{23}\right)^2-y^2}=3$$ $$\sqrt{\frac{841}{529}-y^2}=3-\sqrt{\frac{2704}{529}-y^2}$$ taking squares on both sides, i get $$\sqrt{\frac{2704}{529}-y^2}=\frac{1104}{529}$$ again i take square, $$y^2=\frac{2704}{529}-\left(\frac{1104}{529}\right)^2=\frac{400}{529}$$ $$y=\frac{20}{23}$$

above is the correct value of ordinate of center $C$ of the enclosed circle

Answer 3

In Mr. Michael Rozenberg‘s post, we can easily get the radius $r$ cm of the circle inscribed by the three circles tangent to each other. $$r=\frac{6}{23}$$ First of all, we know that the equations of the circles with radii 1 and 3 are respectively $$ \begin{array}{l} C_1: x^{2}+y^{2}=1 \textrm{ and } C_3: (x-3)^{2}+y^{2}=4 \end{array} $$ Now we dilate the circle (as shown below)$C_1$ and $C_3$ by increasing their radii respectively to $$1+\frac{6}{23} \textrm{ and } 3+\frac{6}{23}, $$

then the new equations can be found as $$ \begin{array}{l} C_1’: x^{2}+y^{2}=\left(1+\frac{6}{23}\right)^2 \textrm{ and } C_3’:x^{2}+(y-4)^{2}=\left(3+\frac{6}{23}\right)^{2} \end{array}\\ $$

Then ordinate of the centre of the inscribed small circle can be easily found by the common chord(horizontal line) of $C_1’$ and $C_3’$.

$$ \begin{aligned} (y-4)^{2}-y^{2} &=\left(\frac{75}{23}\right)^{2}-\left(\frac{29}{23}\right)^{2} \\ y &=\frac{20}{23} \end{aligned} $$

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For completeness, we also dilate $C_2$ to $C_2’$ whose equation is

$$ C_{2}^{\prime}:(x-3)^{2}+y^{2}=\left(2+\frac{6}{23}\right)^{2}. $$ Solving $C_1’$ and $C_2’$ yields $x=\dfrac{21}{23}$ and hence the coordinates of the centre of the inscribed small circle are $\left(\dfrac{21}{23}, \dfrac{20}{23}\right)$.