Question

Inside the circles, $x^2+y^2=1$, there are three smaller circles of equal radius $a$, tangents to each other and to $S$. The value of $a$ equals

• A. $\sqrt{2}(\sqrt{2}-1)$
• B. $\sqrt{3}(2-\sqrt{3})$
• C. $\sqrt{2}(2-\sqrt{3})$
• D. $\sqrt{3}(\sqrt{3}-1)$

Answer 1

The correct option is B $\sqrt{3}(2-\sqrt{3})$

The height of triangle $=\sqrt{(2R{)}^2-R^2}$

$\therefore$ height of triangle $=\sqrt{3}R$

$\therefore$ Now center of mass $=\frac{2}{3}\times$ height of triangle

$\therefore$ Radius $=\frac{2}{3}\sqrt{(2R{)}^2-R^2}+R$

$\therefore$ radius $=\frac{2}{3}\sqrt{3R^2}+R$

$\therefore$ radius $=\frac{2}{\sqrt{3}}R+R$

$\therefore$ radius $=R\left(\frac{2+\sqrt{3}}{\sqrt{3}}\right)$

$f=R\left(\frac{2+\sqrt{3}}{\sqrt{3}}\right)$

$\therefore R=\frac{\sqrt{3}}{(\alpha +\sqrt{3})}$

$\therefore R=\frac{\left(\sqrt{3}\right)(2-\sqrt{3})}{1}$