Question Number 37263 by ajfour last updated on 11/Jun/18
Commented by ajfour last updated on 11/Jun/18
$${If}\:{the}\:{circles}\:\left({radii}\:{given}\right)\:{touch} \\ $$$${in}\:{the}\:{manner}\:{shown}\:{above},\: \\ $$$${Find}\:{coordinates}\:{of}\:{centre}\:{C} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:{and}\:\boldsymbol{{R}}. \\ $$
Answered by MrW3 last updated on 11/Jun/18
![(x_C +a)^2 +y_C ^2 =(R−a)^2 ...(i) (x_C −b)^2 +y_C ^2 =(R−b)^2 ...(ii) (i)−(ii): (x_C +a)^2 −(x_C −b)^2 =(R−a)^2 −(R−b)^2 (a+b)(2x_C +a−b)=(2R−a−b)(−a+b) 2(a+b)x_C +(a+b)(a−b)=−2R(a−b)+(a+b)(a−b) 2(a+b)x_C =−2R(a−b) ⇒x_C =−(((a−b)R)/(a+b)) from (i): y_C ^2 =(R−a)^2 −(x_C +a)^2 =(R+x_C )(R−2a−x_C ) =[R−(((a−b)R)/(a+b))][R−2a+(((a−b)R)/(a+b))] =(((Ra+Rb−Ra+Rb)(Ra+Rb−2a^2 −2ab+Ra−Rb))/((a+b)^2 )) =((4Rab(R−a−b))/((a+b)^2 )) ⇒y_C =±((2(√(ab(R−a−b)R)))/(a+b))](https://www.tinkutara.com/question/Q37315.png)
$$\left({x}_{{C}} +{a}\right)^{\mathrm{2}} +{y}_{{C}} ^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$$\left({x}_{{C}} −{b}\right)^{\mathrm{2}} +{y}_{{C}} ^{\mathrm{2}} =\left({R}−{b}\right)^{\mathrm{2}} \:\:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\left({x}_{{C}} +{a}\right)^{\mathrm{2}} −\left({x}_{{C}} −{b}\right)^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} −\left({R}−{b}\right)^{\mathrm{2}} \\ $$$$\left({a}+{b}\right)\left(\mathrm{2}{x}_{{C}} +{a}−{b}\right)=\left(\mathrm{2}{R}−{a}−{b}\right)\left(−{a}+{b}\right) \\ $$$$\mathrm{2}\left({a}+{b}\right){x}_{{C}} +\left({a}+{b}\right)\left({a}−{b}\right)=−\mathrm{2}{R}\left({a}−{b}\right)+\left({a}+{b}\right)\left({a}−{b}\right) \\ $$$$\mathrm{2}\left({a}+{b}\right){x}_{{C}} =−\mathrm{2}{R}\left({a}−{b}\right) \\ $$$$\Rightarrow{x}_{{C}} =−\frac{\left({a}−{b}\right){R}}{{a}+{b}} \\ $$$$ \\ $$$${from}\:\left({i}\right): \\ $$$${y}_{{C}} ^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} −\left({x}_{{C}} +{a}\right)^{\mathrm{2}} =\left({R}+{x}_{{C}} \right)\left({R}−\mathrm{2}{a}−{x}_{{C}} \right) \\ $$$$=\left[{R}−\frac{\left({a}−{b}\right){R}}{{a}+{b}}\right]\left[{R}−\mathrm{2}{a}+\frac{\left({a}−{b}\right){R}}{{a}+{b}}\right] \\ $$$$=\frac{\left({Ra}+{Rb}−{Ra}+{Rb}\right)\left({Ra}+{Rb}−\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{ab}+{Ra}−{Rb}\right)}{\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{Rab}\left({R}−{a}−{b}\right)}{\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{y}_{{C}} =\pm\frac{\mathrm{2}\sqrt{{ab}\left({R}−{a}−{b}\right){R}}}{{a}+{b}} \\ $$
Commented by ajfour last updated on 11/Jun/18
$${Excellent}\:{again}\:{sir}. \\ $$$${y}_{{C}} =\mathrm{0}\:\:{if}\:{a}+{b}={R}\:.\:{Fine}\:{enough}. \\ $$