Question Number 214100 by ajfour last updated on 28/Nov/24
Answered by mr W last updated on 28/Nov/24
Commented by mr W last updated on 28/Nov/24
![for maximum small circle: AF⊥FD FE=(√((a+r)^2 −(a−r)^2 ))=2(√(ar)) (a+b)CD^2 +b(a+r)^2 =(a+2b)[(b+r)^2 +(a+b)b] CD^2 =(((a+2b)[(b+r)^2 +(a+b)b]−b(a+r)^2 )/(a+b)) ED=(√((((a+2b)[(b+r)^2 +(a+b)b]−b(a+r)^2 )/(a+b))−r^2 )) FD=2(√(ar))+(√((((a+2b)[(b+r)^2 +(a+b)b]−b(a+r)^2 )/(a+b))−r^2 )) FD^2 =a^2 +(a+2b)^2 ⇒{2(√(ar))+(√((((a+2b)[(b+r)^2 +(a+b)b]−b(a+r)^2 )/(a+b))−r^2 ))}^2 =a^2 +(a+2b)^2](https://www.tinkutara.com/question/Q214121.png)
$${for}\:{maximum}\:{small}\:{circle}: \\ $$$${AF}\bot{FD} \\ $$$${FE}=\sqrt{\left({a}+{r}\right)^{\mathrm{2}} −\left({a}−{r}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{ar}} \\ $$$$\left({a}+{b}\right){CD}^{\mathrm{2}} +{b}\left({a}+{r}\right)^{\mathrm{2}} =\left({a}+\mathrm{2}{b}\right)\left[\left({b}+{r}\right)^{\mathrm{2}} +\left({a}+{b}\right){b}\right] \\ $$$${CD}^{\mathrm{2}} =\frac{\left({a}+\mathrm{2}{b}\right)\left[\left({b}+{r}\right)^{\mathrm{2}} +\left({a}+{b}\right){b}\right]−{b}\left({a}+{r}\right)^{\mathrm{2}} }{{a}+{b}} \\ $$$${ED}=\sqrt{\frac{\left({a}+\mathrm{2}{b}\right)\left[\left({b}+{r}\right)^{\mathrm{2}} +\left({a}+{b}\right){b}\right]−{b}\left({a}+{r}\right)^{\mathrm{2}} }{{a}+{b}}−{r}^{\mathrm{2}} } \\ $$$${FD}=\mathrm{2}\sqrt{{ar}}+\sqrt{\frac{\left({a}+\mathrm{2}{b}\right)\left[\left({b}+{r}\right)^{\mathrm{2}} +\left({a}+{b}\right){b}\right]−{b}\left({a}+{r}\right)^{\mathrm{2}} }{{a}+{b}}−{r}^{\mathrm{2}} } \\ $$$${FD}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left({a}+\mathrm{2}{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left\{\mathrm{2}\sqrt{{ar}}+\sqrt{\frac{\left({a}+\mathrm{2}{b}\right)\left[\left({b}+{r}\right)^{\mathrm{2}} +\left({a}+{b}\right){b}\right]−{b}\left({a}+{r}\right)^{\mathrm{2}} }{{a}+{b}}−{r}^{\mathrm{2}} }\right\}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left({a}+\mathrm{2}{b}\right)^{\mathrm{2}} \\ $$