Question

The centre O of a circle of radius 2 cm lies on the origin. Another circle C(O′, r) touches the given circle at A, positive X-axis and positive Y-axis at B and C respectively. The length of their direct common tangent which has positive slope is

• A.
  $16(\sqrt{2}+1)cm$
• B.
  $4(1+\sqrt{2})cm$
• C.
  $2\sqrt{1+\sqrt{2}}cm$
• D.
  $4\sqrt{\sqrt{2}+1}cm$

Answer 1

The correct option is D

$4\sqrt{\sqrt{2}+1}cm$
Given that a circle with centre at O and radius is 2 cm.




Let us consider another circle with centre O′ and radius r.

In $\Delta O\text{'}BO,OO\text{'}=OA+AO\text{'}=2+rHere,OCO\text{'}Bisasquareofsider.\therefore O\text{'}B=OB=rIn\Delta O\text{'}BO,byPythagorastheorem,\Rightarrow (OO\text{'}{)}^2=O{B}^2+(O\text{'}B{)}^2\Rightarrow (r+2{)}^2=r^2+r^2\Rightarrow r^2+4r+4=2r^2\Rightarrow r^2–4r–4=0\Rightarrow r=\frac{-(-4)\pm \sqrt{(-4{)}^2-4\times 1\times (-4)}}{2}\Rightarrow r=\frac{4\pm \sqrt{16+16}}{2}\Rightarrow r=\frac{4\pm 4\sqrt{2}}{2}=2\pm 2\sqrt{2}$

Since, r cannot be negative.

Thus, $r=2+2\sqrt{2}$

Now, $O\text{'}D=O\text{'}F–FD=r–OE(\because ODFEisasquare)=2+2\sqrt{2}-2=2\sqrt{2}$

Thus, $O\text{'}Dis2\sqrt{2}$ cm.

Also, in $\Delta OO\text{'}D$, by Pythagoras theorem,

$(OO\text{'}{)}^2=(OD{)}^2+(O^{\prime }D{)}^2\Rightarrow OD=\sqrt{(O{O}^{\prime }{)}^2-(O^{\prime }D{)}^2}\Rightarrow OD=\sqrt{{(4+2\sqrt{2})}^2-{\left(2\sqrt{2}\right)}^2}(\because O{O}^{\prime }=r+2)\Rightarrow OD=\sqrt{4^2+(2\sqrt{2}{)}^2+2\times 4\times 2\sqrt{2}-(2\sqrt{2}{)}^2}\Rightarrow OD=4\sqrt{\sqrt{2}+1}cm$

Now, OD = EF (By construction)

Thus, the length of common tangent is$4\sqrt{\sqrt{2}+1}$ cm.
Hence the correct answer is option d.