Question Number 215147 by AlagaIbile last updated on 30/Dec/24
Answered by A5T last updated on 30/Dec/24
Commented by A5T last updated on 30/Dec/24
$$\mathrm{OP}=\sqrt{\left(\mathrm{4}.\mathrm{5}−\mathrm{r}\right)^{\mathrm{2}} −\left(\mathrm{2}.\mathrm{5}−\mathrm{r}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\mathrm{7}−\mathrm{2r}\right)} \\ $$$$\Rightarrow\mathrm{GO}=\sqrt{\mathrm{OP}^{\mathrm{2}} +\mathrm{GP}^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\mathrm{7}−\mathrm{2r}\right)+\left(\mathrm{2}+\mathrm{r}\right)^{\mathrm{2}} }=\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{18}} \\ $$$$\Rightarrow\mathrm{GC}=\sqrt{\mathrm{GO}^{\mathrm{2}} −\mathrm{OC}^{\mathrm{2}} }=\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{18}−\mathrm{r}^{\mathrm{2}} }=\sqrt{\mathrm{18}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$
Answered by mr W last updated on 30/Dec/24
Commented by mr W last updated on 30/Dec/24
$${let}'{s}\:{look}\:{at}\:{the}\:{general}\:{case}. \\ $$$${R}=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${OB}={R}−{a}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$${GA}^{\mathrm{2}} =\left(\frac{{a}+{b}}{\mathrm{2}}−{r}\right)^{\mathrm{2}} −\left(\frac{{b}−{a}}{\mathrm{2}}−{r}\right)^{\mathrm{2}} +\left({a}+{r}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:={a}\left({a}+{b}\right)+{r}^{\mathrm{2}} \\ $$$${GC}^{\mathrm{2}} ={GA}^{\mathrm{2}} −{r}^{\mathrm{2}} ={a}\left({a}+{b}\right) \\ $$$$\Rightarrow{GC}=\sqrt{{a}\left({a}+{b}\right)}=\sqrt{\mathrm{2}×\left(\mathrm{2}+\mathrm{7}\right)}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$
Commented by mr W last updated on 30/Dec/24
$${GC}\:{is}\:{independent}\:{from}\:{the}\:{size}\:{of} \\ $$$${the}\:{small}\:{circle}. \\ $$
Commented by mr W last updated on 30/Dec/24
Commented by ajfour last updated on 30/Dec/24
https://youtu.be/qdhFOxn2SzU?si=QwAhflQzh-f_rJ8L
my first video lecture on Inductance