Question Number 15480 by ajfour last updated on 10/Jun/17
Commented by ajfour last updated on 10/Jun/17
$${Q}.\:\mathrm{15434}\:\:{answered}\:{here}. \\ $$$$\:{To}\:{find}\:{maximum}\:{length}\:{of} \\ $$$${AB},\:{in}\:{terms}\:{of}\:\boldsymbol{{r}},\boldsymbol{{R}},\:{and}\:\boldsymbol{{d}}\:. \\ $$
Answered by ajfour last updated on 11/Jun/17
![To find P(h,k). x^2 +y^2 =r^2 (x−d)^2 +y^2 =R^2 subtracting to obtain h 2dh=R^2 −r^2 h=((R^2 −r^2 )/(2d)) k= (√(r^2 −h^2 )) (considering just the positive one here) let equation of APB be, y=m(x−h)+k solving this with x^2 +y^2 =r^2 x^2 +[m(x−h)+k]^2 −r^2 =0 (1+m^2 )x^2 +(−2m^2 h+2mk)x+c_1 =0 its roots are h and x_1 . so x_1 +h=−(((−2m^2 h+2mk)/(1+m^2 ))) ...(i) Again solving line y=m(x−h)+k witb (x−d)^2 +y^2 =R^2 (x−d)^2 +[m(x−h)+k]^2 −R^2 =0 (1+m^2 )x^2 +(−2d−2m^2 h+2mk)x+c_2 =0 its roots are h, and x_2 . x_2 +h=−(((−2d−2m^2 h+2mk)/(1+m^2 ))) ..(ii) eqn. (i)−eqn.(ii) gives: x_2 −x_1 =((2d)/(1+m^2 )) length of AB = (√(1+m^2 ))(x_2 −x_1 ) l_(AB) = ((2d)/( (√(1+m^2 )))) and is clearly maximum for m=0, and (length AB)_(max) = 2d .](https://www.tinkutara.com/question/Q15482.png)
$$\:{To}\:{find}\:{P}\left({h},{k}\right). \\ $$$$\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\:\left({x}−{d}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${subtracting}\:{to}\:\:{obtain}\:\boldsymbol{{h}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}{dh}={R}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\:\:{h}=\frac{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{d}} \\ $$$$\:\:\:{k}=\:\:\sqrt{{r}^{\mathrm{2}} −{h}^{\mathrm{2}} }\:\:\:\:\left({considering}\:{just}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{the}\:{positive}\:{one}\:{here}\right) \\ $$$${let}\:{equation}\:{of}\:{APB}\:{be}, \\ $$$$\:\:\:\:\:\:\:\:{y}={m}\left({x}−{h}\right)+{k} \\ $$$${solving}\:{this}\:{with}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{2}} +\left[{m}\left({x}−{h}\right)+{k}\right]^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{1}+{m}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\left(−\mathrm{2}{m}^{\mathrm{2}} {h}+\mathrm{2}{mk}\right){x}+{c}_{\mathrm{1}} =\mathrm{0} \\ $$$${its}\:{roots}\:{are}\:\:{h}\:{and}\:{x}_{\mathrm{1}} . \\ $$$${so}\:\:\:\:{x}_{\mathrm{1}} +{h}=−\left(\frac{−\mathrm{2}{m}^{\mathrm{2}} {h}+\mathrm{2}{mk}}{\mathrm{1}+{m}^{\mathrm{2}} }\right)\:\:\:\:…\left({i}\right) \\ $$$${Again}\:{solving}\:{line} \\ $$$$\:\:\:{y}={m}\left({x}−{h}\right)+{k} \\ $$$$\:{witb}\:\left({x}−{d}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\:\:\left({x}−{d}\right)^{\mathrm{2}} +\left[{m}\left({x}−{h}\right)+{k}\right]^{\mathrm{2}} −{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\left(\mathrm{1}+{m}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\left(−\mathrm{2}{d}−\mathrm{2}{m}^{\mathrm{2}} {h}+\mathrm{2}{mk}\right){x}+{c}_{\mathrm{2}} =\mathrm{0} \\ $$$${its}\:{roots}\:{are}\:\boldsymbol{{h}},\:{and}\:\boldsymbol{{x}}_{\mathrm{2}} . \\ $$$$\:\:{x}_{\mathrm{2}} +{h}=−\left(\frac{−\mathrm{2}{d}−\mathrm{2}{m}^{\mathrm{2}} {h}+\mathrm{2}{mk}}{\mathrm{1}+{m}^{\mathrm{2}} }\right)\:\:\:..\left({ii}\right) \\ $$$${eqn}.\:\left({i}\right)−{eqn}.\left({ii}\right)\:{gives}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}_{\mathrm{2}} −{x}_{\mathrm{1}} =\frac{\mathrm{2}{d}}{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$${length}\:{of}\:{AB}\:\:=\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{l}_{{AB}} =\:\frac{\mathrm{2}{d}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\:{and}\:{is}\:{clearly} \\ $$$${maximum}\:{for}\:{m}=\mathrm{0},\:{and} \\ $$$$\:\left({length}\:{AB}\right)_{{max}} =\:\mathrm{2}{d}\:. \\ $$$$ \\ $$
Commented by mrW1 last updated on 11/Jun/17
$$\boldsymbol{\mathcal{V}}\mathcal{ERY}\:\mathcal{NICE} \\ $$