Question Number 212784 by Spillover last updated on 23/Oct/24
Answered by A5T last updated on 24/Oct/24
$${d}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}\left({r}−{a}\right){cos}\theta \\ $$$${c}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{r}\left({r}−{a}\right){cos}\theta \\ $$$$\Rightarrow{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2}\left({r}−{a}\right)^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} −\mathrm{4}{ar} \\ $$$${b}=\left(\mathrm{2}{r}−{a}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{4}{ar}=\mathrm{4}{r}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} −\mathrm{4}{ar} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$
Commented by Spillover last updated on 24/Oct/24
$${thank}\:{you} \\ $$