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There are three circles of radii $5 cm$, $9cm$ & $11 cm$ touching each other externally. What will be the radius of the largest circle inscribed in the region bounded by three circles? Thus inscribed circles is touching three circles at three different points.

I know that the radius $r$ of the largest circle inscribed in the region bounded by three externally touching identical circles each with a radius $R$ is given as $$r=R\left(\frac{2}{\sqrt{3}}-1\right)$$

I have studied maths up to 12th. Thanks!

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    $\begingroup$ This is the Problem of Apollonius, solved by Decartes' theorem and commemorated in Soddy's "Kiss precise", some of which is quoted in the first article. $\endgroup$ – Ross Millikan Jun 8 '15 at 14:26
  • $\begingroup$ yes, but kindly explain it? $\endgroup$ – Bhaskara-III Jun 8 '15 at 14:27
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    $\begingroup$ The theorem gives a simple formula for the inverse radius of the fourth circle, given the inverse radii of the other three. "The sum of the squares of all the bends is half the square of their sum" There are two solutions, one for the small circle you seek and on for a large circle that goes around your three starting circles. $\endgroup$ – Ross Millikan Jun 8 '15 at 14:30
  • $\begingroup$ is it 10.1 cm? plz $\endgroup$ – RE60K Jun 8 '15 at 14:33
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    $\begingroup$ To me this looks like the given three circles touch only the solution circle: Problem of Apollonius $\endgroup$ – mvw Jun 8 '15 at 14:50
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$2$ circles touch externally $\iff$ sum of their radii $=$ distance between their centers. Taking unit length to be $1$cm in Cartesian coordinate system we can express the given circles in the following way $$x^2+y^2=11^2\cdots(1)\\x^2+(y-20)^2=9^2\cdots(2)\\(x-a)^2+(y-b)^2=5^2\cdots(3)$$ As $(3)$ touches $(1),(2)$ externally, $$a^2+b^2=(5+11)^2,a^2+(b-20)^2=(5+9)^2\Rightarrow a={3\sqrt{55}\over 2},b={23\over 2}$$ The largest circle, let's call it $(4)$, inside the region bounded by $(1),(2),(3)$, it touches $(1),(2),(3)$. Suppose $(4)$ is given by $$(x-c)^2+(y-d)^2=r^2\\\therefore c^2+d^2=(r+11)^2\cdots(5)\\c^2+(d-20)^2=(r+9)^2\cdots(6)\\(c-a)^2+(d-b)^2=(r+5)^2\cdots(7)\\ (6)-(5)\equiv d={r\over 10}+11\cdots(8)\\(5)-(8)^2\equiv c^2=99{r\over 10}\left({r\over 10}+2\right)\cdots(9)$$ Let ${r\over 10}=q$. $$(7)-(5)\equiv ac+bd=60q+176\Rightarrow c={97q+99\over 2a}\cdots(10)$$ $(10)^2-(9)$ gives us a quadratic equation in $q$ solving which we get $$r={495(-199+30\sqrt{55})\over 9899}\approx1.17$$enter image description here

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  • $\begingroup$ +1 Personally, I prefer solving this using Descartes theorem. i.e $r$ is the positive root for $$\left(\frac{1}{r} + \frac{1}{5} + \frac{1}{9} + \frac{1}{11}\right)^2 = 2\left(\frac{1}{r^2} + \frac{1}{5^2} + \frac{1}{9^2} + \frac{1}{11^2}\right)$$ $\endgroup$ – achille hui Jun 8 '15 at 17:21
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    $\begingroup$ He said the $12$-th grade thing so I thought this would be easier : ) $\endgroup$ – Jack's wasted life Jun 8 '15 at 17:24
  • $\begingroup$ thanks for your help $\endgroup$ – Bhaskara-III Jun 16 '15 at 0:50

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