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Question Number 135712 by Dwaipayan Shikari last updated on 15/Mar/21

Commented by Dwaipayan Shikari last updated on 15/Mar/21

Smaller circle has half of the radius of the larger circle. Prove that, irrespective of any position oc the smaller circle that blue point will always stay on the diametre of that larger circle

$${Smaller}\:{circle}\:{has}\:{half}\:{of}\:{the}\:{radius}\:{of}\:{the}\:{larger}\:{circle}. \\ $$$${Prove}\:{that},\:{irrespective}\:{of}\:{any}\:{position}\:{oc}\:{the}\:{smaller}\:{circle}\: \\ $$$${that}\:{blue}\:{point}\:{will}\:{always}\:{stay}\:{on}\:{the}\:{diametre}\:{of}\:{that}\: \\ $$$${larger}\:{circle} \\ $$

Answered by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

x_C =(R−r)sin θ y_C =−(R−r)cos θ Rθ=rϕ ⇒ϕ=(R/r)θ x_P =x_C −r sin (ϕ−θ) ⇒x_P =(R−r)sin θ−r sin ((R/r)−1)θ y_P =y_C −r cos (ϕ−θ) ⇒y_P =−(R−r)cos θ−r cos ((R/r)−1)θ the locus of P see diagrams. special case R=2r: x_P =rsin θ−r sin θ=0 y_P =−r cos θ−r cos θ=−R cos θ that means P is always on the diameter of the larger circle x=0, y=−R to R.

$${x}_{{C}} =\left({R}−{r}\right)\mathrm{sin}\:\theta \\ $$$${y}_{{C}} =−\left({R}−{r}\right)\mathrm{cos}\:\theta \\ $$$${R}\theta={r}\varphi \\ $$$$\Rightarrow\varphi=\frac{{R}}{{r}}\theta \\ $$$${x}_{{P}} ={x}_{{C}} −{r}\:\mathrm{sin}\:\left(\varphi−\theta\right) \\ $$$$\Rightarrow{x}_{{P}} =\left({R}−{r}\right)\mathrm{sin}\:\theta−{r}\:\mathrm{sin}\:\left(\frac{{R}}{{r}}−\mathrm{1}\right)\theta \\ $$$${y}_{{P}} ={y}_{{C}} −{r}\:\mathrm{cos}\:\left(\varphi−\theta\right) \\ $$$$\Rightarrow{y}_{{P}} =−\left({R}−{r}\right)\mathrm{cos}\:\theta−{r}\:\mathrm{cos}\:\left(\frac{{R}}{{r}}−\mathrm{1}\right)\theta \\ $$$${the}\:{locus}\:{of}\:{P}\:{see}\:{diagrams}. \\ $$$$ \\ $$$$\boldsymbol{{special}}\:\boldsymbol{{case}}\:\boldsymbol{{R}}=\mathrm{2}\boldsymbol{{r}}: \\ $$$${x}_{{P}} ={r}\mathrm{sin}\:\theta−{r}\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$${y}_{{P}} =−{r}\:\mathrm{cos}\:\theta−{r}\:\mathrm{cos}\:\theta=−{R}\:\mathrm{cos}\:\theta \\ $$$${that}\:{means}\:{P}\:{is}\:{always}\:{on}\:{the} \\ $$$${diameter}\:{of}\:{the}\:{larger}\:{circle} \\ $$$${x}=\mathrm{0},\:{y}=−{R}\:{to}\:{R}. \\ $$

Commented by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

Commented by Dwaipayan Shikari last updated on 15/Mar/21

Which application you are using sir? Thanks sir ! Great sir!

$${Which}\:{application}\:{you}\:{are}\:{using}\:{sir}? \\ $$$${Thanks}\:{sir}\:!\:{Great}\:{sir}! \\ $$

Commented by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

i used Grapher as always.

$${i}\:{used}\:{Grapher}\:{as}\:{always}. \\ $$

Commented by liberty last updated on 15/Mar/21

love grapher

$${love}\:{grapher} \\ $$

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