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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.

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  • $\begingroup$ i have given a nice answer to you. u can check it will be correct value. $\endgroup$ – Bhaskara-III Aug 9 '16 at 14:32
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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.

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  • $\begingroup$ can you show how many steps does it take to reach the answer please ? because i have solved by different method $\endgroup$ – Bhaskara-III Sep 11 '16 at 10:38
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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

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  • $\begingroup$ I will give you +1 for your effort...but you know I believed there would be a simpler solution. The answer I chose has less numerical calculations see. Anyways keep up the hard work $\endgroup$ – User Not Found Sep 8 '16 at 10:18
  • $\begingroup$ i give you correct answer. how many steps does it take to solve by the method used in the above answer? i think that is too lengthy too. $\endgroup$ – Bhaskara-III Sep 11 '16 at 10:11
  • $\begingroup$ See I acknowledge your work but when trying to solve it your method it will take some time to make the computations but the answer I accepted would just mean entering the equations in a computer and getting the answer...better than entering numbers. Another thing I didn't want to say is that it is bad to accept an answer and then change to another answer...that would make me a man with no honour. So I can't do that. $\endgroup$ – User Not Found Sep 11 '16 at 18:21
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joining the radius will of the circles C1,C2,C3 will form a right angled triangle,rest can be done by DISTANCE formula, you will get the equations. Solve them and get your answer. THANKS

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