Question

The figure below shows circles with diameter AB, BC, CD and AD so that the line EFG is the common tangent to circles with diameter AB, BC and CD at E, F and G, respectively. If $EF=a$ and $FG=b$, prove that the shaded area $S=\frac{\pi }{4}(a^2+b^2+ab)$

Answer 1

Let the radius of circles AB, BC, CD, and AD are $r_1,r_2,r_3,$ and $r$ respectively.

Given that the line EFG is the common tangent to circles AB, BC and CD at E, F and G, respectively.

We know that if two circles are touching outside then the length $l$ of the common tangent is given by :

$l^2=4c_1{c}_2$

where $c_1$ and $c_2$ are the radius of the circles.

$\therefore 4r_1{r}_2=a^2\dots \dots \left(1\right)4r_2{r}_3=b^2\dots \dots \left(2\right)4r_2^2+4r_1{r}_2+4r_2{r}_3+4r_1{r}_3=(a+b{)}^2$

$\Rightarrow 4r_2^2+a^2+b^2+4r_1{r}_3=a^2+b^2+2ab$

$\Rightarrow 2r_2^2+2r_1{r}_3=ab\dots \dots \left(3\right)$

From 1 and 2

$ab=4r_2\sqrt{r_1{r}_2}\dots \dots \left(4\right)$

From 3 and 4

$4r_2\sqrt{r_1{r}_3}=2r_2^2+2r_1{r}_3$

$(r_2-\sqrt{r_1{r}_3}{)}^2=0r_2=\sqrt{r_1{r}_3}$

Therefore, from 4

$4r_1{r}_2=ab$

$\therefore$$S=\frac{\pi }{2}\left(\right(r^2-r_1^2-r_2^2-r_3^2)$

and we know that $2r=2r_1+2r_2+2r_3$

$S=\frac{\pi }{2}\left(\right(r_1+r_2+r_3{)}^2-r_1^2-r_2^2-r_3^2)$

$S=\frac{\pi }{2}(2r_1{r}_2+2r_2{r}_3+2r_3{r}_4)$

$S=\frac{\pi }{4}(a^2+b^2+ab)$