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!
Best Answer
$2$ 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 $$r={495(-199+30\sqrt{55})\over 9899}\approx1.17$$