Question

In the given figure, two circles of different radii are placed against a right angle. If the radius of the bigger circle is 1 unit, then the radius of the smaller circle is

• A.
  $3-2\sqrt{2}units$
• B.
  $6-4\sqrt{2}units$
• C.
  $5-2\sqrt{2}units$
• D.
  $3-4\sqrt{2}units$

Answer 1

The correct option is A

$3-2\sqrt{2}units$
Let the radii of the bigger and smaller circles be R and r, respectively.






In the right triangles OBC' and OAC, by Pythagoras theorem,

$O{C}^{\prime }=\sqrt{(C^{\prime }B{)}^2+(OB{)}^2}=\sqrt{R^2+R^2}=R\sqrt{2}Also,OC=\sqrt{(CA{)}^2+(OA{)}^2}=\sqrt{r^2+r^2}=r\sqrt{2}$

Now,
$OC\text{'}=C\text{'}D+DC+CO$
$\Rightarrow R\sqrt{2}=R+r+r\sqrt{2}\Rightarrow r(1+\sqrt{2})=R(\sqrt{2}-1)\Rightarrow r=\frac{R(\sqrt{2}-1)}{(1+\sqrt{2})}\Rightarrow r=\frac{R(\sqrt{2}-1)}{(\sqrt{2}+1)}\times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}$


It is given that the radius of the bigger circle is 1 unit,

i.e., R = 1 unit.
$\therefore r=(3-2\sqrt{2}units)$
Hence, the correct answer is option a.