Q. 49
Question
A circle is inscribed in a square so that each side of the square is tangent to the circle. A smaller circle is inscribed in the square so that this circle is tangent to two sides of the square and is tangent to the larger circle, as shown in the following figure:
What is the ratio of the radius of the smaller circle to the radius of the larger circle?
Step-by-Step Solution
VerifiedThe ratio of the radius of the smaller circle to the radius of the larger circle is
\(3-2\sqrt{2}\)
Let the side of the square be 'a' unit.
therefore the radius of the larger circle becomes \(\frac{a}{2}\) from the figure as all four sides of the square are tangents to the larger circle.
another smaller circle is inscribed in such a way that it touches two adjacent sides of the square and the larger circle.
Considering triangle OAB and applying Pythagoras theorem,
\(OB^2= OA^2 + AB^2\)
Where, OA = \(\frac{a}{2}\)
AB = \(\frac{a}{2}\)
\( OB^2 = {a^2}/4+{a^2}/4\)
OB = \(\frac{a}{{\sqrt 2 }}\)
From the figure,
BC = OB-OC
= \(\frac{a}{{\sqrt 2 }} - \frac{a}{2}\)
= \(\frac{{a\left( {\sqrt 2 - 1} \right)}}{2}\)
Let, the radius of the smaller circle be y and the length BD = z
applying Pythagoras theorem to the triangle DBE:
\(BD^2 = DE^2+ EB^2\) (DE = y, EB = y)
\( z^2 =y^2+y^2\)
z =\(\sqrt 2 y\)
BD =z =\(\sqrt 2 y\)
From the figure,
BD = BC-CD
\(\sqrt 2 y\) = \(\frac{{a\left( {\sqrt 2 - 1} \right)}}{2}-y\)
\(\frac{y}{{\frac{a}{2}}} = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}\)
\(\frac{y}{{\frac{a}{2}}}\)= ratio of the radius of the smaller circle to that of the larger circle
rationalizing the above fraction:
\(\frac{y}{{\frac{a}{2}}} = \frac{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}}\)
solving, we get the ratio = \(3-2\sqrt{2}\)