Question

A circle is drawn in a sector of a larger circle of radius r, as shown in the adjacent figure. The smallest circle is tangent to the two bounding radii and the arc of the sector. The radius of the small circle is

• A. $\frac{r}{2}$
• B. $\frac{r}{3}$
• C. $\frac{2\sqrt{3}r}{5}$
• D. $\frac{r}{\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{r}{3}$

Let the center of the smaller circle be A, center of the bigger circle be O and one of the point of contacts of the tangents to the smaller circle be B.

The point where the arc is tangent to the smaller circle be C.

In $\Delta AOB$, let $AO=x$.

Since $\angle AOB={60}^o,AB=\frac{x}{2}$ = radius of the smaller circle.

Also, $AC=\frac{x}{2}$

$\therefore AO+AC=OC=r$

$\Rightarrow x+\frac{x}{2}=r$

$\Rightarrow \frac{3x}{2}=r$ and $\frac{x}{2}=\frac{r}{3}$