In the accompanying diagram, a circle of radius $r$ is tangent to both sides of the right-angled corner. What is the radius of the largest circle that will fit in the same corner between the larger circle and the corner?
[Math] Find the radius of the largest circle
algebra-precalculuscalculuscircles
Related Solutions
$2$ 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 $$r={495(-199+30\sqrt{55})\over 9899}\approx1.17$$
Drop a perpendicular from the center of the circle to one of the sides and draw the radius to one of its endpoints. This should give you some nice triangles to play with :)
Best Answer
I think this image makes this problem much easier. By the Pythagorean Theorem we have $$r^2 + r^2 = (x\sqrt 2 + x + r)^2$$
Solving for $x$ we get:
$$r\sqrt 2 = x(\sqrt 2 + 1) + r$$
$$r(\sqrt 2 - 1) = x(\sqrt 2 + 1)$$
$$x = \frac{\sqrt 2 - 1}{\sqrt 2 + 1}r=\frac{(\sqrt 2 - 1)(\sqrt 2 - 1)}{(\sqrt 2 + 1)(\sqrt 2 - 1)}r=\frac{2-2\sqrt 2 + 1}{1}r=(3-2\sqrt 2)r$$