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Question Number 52198 by ajfour last updated on 04/Jan/19

Answered by MJS last updated on 04/Jan/19

c_1 : x^2 +y^2 =4R^2   c_2 : x^2 +(y−R)^2 =R^2   c_3 : (x−q)^2 +(y−r)^2 =r^2   now intersect c_1 ∩c_3  and c_2 ∩c_3  with exactly  one solution each  ⇒ q=(√2)R, r=(1/2)R

$${c}_{\mathrm{1}} :\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} \\ $$$${c}_{\mathrm{2}} :\:{x}^{\mathrm{2}} +\left({y}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${c}_{\mathrm{3}} :\:\left({x}−{q}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{intersect}\:{c}_{\mathrm{1}} \cap{c}_{\mathrm{3}} \:\mathrm{and}\:{c}_{\mathrm{2}} \cap{c}_{\mathrm{3}} \:\mathrm{with}\:\mathrm{exactly} \\ $$$$\mathrm{one}\:\mathrm{solution}\:\mathrm{each} \\ $$$$\Rightarrow\:{q}=\sqrt{\mathrm{2}}{R},\:{r}=\frac{\mathrm{1}}{\mathrm{2}}{R} \\ $$

Commented by ajfour last updated on 04/Jan/19

okay Sir, thanks for Answer, i  shall try solving.

$${okay}\:{Sir},\:{thanks}\:{for}\:{Answer},\:{i} \\ $$$${shall}\:{try}\:{solving}. \\ $$

Answered by mr W last updated on 04/Jan/19

Commented by ajfour last updated on 04/Jan/19

Wonderful Sir!

$${Wonderful}\:{Sir}! \\ $$

Commented by mr W last updated on 04/Jan/19

(R+r)^2 −(R−r)^2 =(2R−r)^2 −r^2   4Rr=4R(R−r)  ⇒(R/r)=2

$$\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} =\left(\mathrm{2}{R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\mathrm{4}{Rr}=\mathrm{4}{R}\left({R}−{r}\right) \\ $$$$\Rightarrow\frac{{R}}{{r}}=\mathrm{2} \\ $$

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