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Question Number 46338 by ajfour last updated on 24/Oct/18

Commented by MJS last updated on 24/Oct/18

is C the center of the small circle?

$$\mathrm{is}\:{C}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circle}? \\ $$

Commented by ajfour last updated on 24/Oct/18

yes Sir.

$${yes}\:{Sir}. \\ $$

Answered by MJS last updated on 24/Oct/18

I put x=r to not get confused    please check. the path is ok but I′ve been in  a hurry so maybe I made some mistakes    A= (((r−1)),(0) )  C= (((−1)),(0) )  D=circle_1 ∩circle_2   (1)         x^2 +y^2 =1  (2)         (x+1)^2 +y^2 =r^2   (2−1)  2x+1=r^2 −1 ⇒ x=((r^2 −2)/2) ⇒ y=−((r(√(4−r^2 )))/2)  D= ((((r^2 −2)/2)),((−((r(√(4−r^2 )))/2))) )  ∣Ol_(AD) ∣=r  l_(AD) : r(√(4−r^2 ))x+r(r−2)y−r(r−1)(√(4−r^2 ))=0  ∣Ol_(AD) ∣=(((r−1)(√(r+2)))/2)=r  r^3 −4r^2 −3r+2=0  r_1 =−1; r_2 =(5/2)−((√(17))/2); r_3 =(5/2)+((√(17))/2)  ⇒ r=(5/2)−((√(17))/2)≈.438447

$$\mathrm{I}\:\mathrm{put}\:{x}={r}\:\mathrm{to}\:\mathrm{not}\:\mathrm{get}\:\mathrm{confused} \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{check}.\:\mathrm{the}\:\mathrm{path}\:\mathrm{is}\:\mathrm{ok}\:\mathrm{but}\:\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{in} \\ $$$$\mathrm{a}\:\mathrm{hurry}\:\mathrm{so}\:\mathrm{maybe}\:\mathrm{I}\:\mathrm{made}\:\mathrm{some}\:\mathrm{mistakes} \\ $$$$ \\ $$$${A}=\begin{pmatrix}{{r}−\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{−\mathrm{1}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${D}=\mathrm{circle}_{\mathrm{1}} \cap\mathrm{circle}_{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}−\mathrm{1}\right)\:\:\mathrm{2}{x}+\mathrm{1}={r}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow\:{x}=\frac{{r}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}}\:\Rightarrow\:{y}=−\frac{{r}\sqrt{\mathrm{4}−{r}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${D}=\begin{pmatrix}{\frac{{r}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}}}\\{−\frac{{r}\sqrt{\mathrm{4}−{r}^{\mathrm{2}} }}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mid{Ol}_{{AD}} \mid={r} \\ $$$${l}_{{AD}} :\:{r}\sqrt{\mathrm{4}−{r}^{\mathrm{2}} }{x}+{r}\left({r}−\mathrm{2}\right){y}−{r}\left({r}−\mathrm{1}\right)\sqrt{\mathrm{4}−{r}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mid{Ol}_{{AD}} \mid=\frac{\left({r}−\mathrm{1}\right)\sqrt{{r}+\mathrm{2}}}{\mathrm{2}}={r} \\ $$$${r}^{\mathrm{3}} −\mathrm{4}{r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{2}=\mathrm{0} \\ $$$${r}_{\mathrm{1}} =−\mathrm{1};\:{r}_{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{17}}}{\mathrm{2}};\:{r}_{\mathrm{3}} =\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{r}=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\approx.\mathrm{438447} \\ $$

Commented by MJS last updated on 24/Oct/18

ok thank you.

$$\mathrm{ok}\:\mathrm{thank}\:\mathrm{you}. \\ $$

Commented by ajfour last updated on 24/Oct/18

I had got the same answer Sir.

$${I}\:{had}\:{got}\:{the}\:{same}\:{answer}\:{Sir}. \\ $$

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