Question Number 61559 by Tawa1 last updated on 04/Jun/19
Commented by Tawa1 last updated on 04/Jun/19
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pink} \\ $$
Answered by ajfour last updated on 04/Jun/19
![x^2 +y^2 =2 (right circle) (x+(√2)+1)^2 +(y−1)^2 =2 (left one) ⇒ 2((√2)+1)x+3+2(√2)−2y+1=0 ⇒ y=((√2)+1)x+(√2)+2 ⇒ x^2 +[((√2)+1)x+(√2)+2]^2 =2 ⇒ (4+2(√2))x^2 +2(√2)(3+2(√2))x+4(√2)=0 one root is x=−(√2) ⇒ other root =−((2(√2)((√2)+1)^2 )/(2(√2)((√2)+1)))+(√2) = −(√2)−1+(√2)= −1 y=1 pink segment length=2](https://www.tinkutara.com/question/Q61561.png)
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}\:\:\:\:\left({right}\:{circle}\right) \\ $$$$\left({x}+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}\:\:\:\left({left}\:{one}\right) \\ $$$$\Rightarrow\:\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){x}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:{y}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){x}+\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +\left[\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){x}+\sqrt{\mathrm{2}}+\mathrm{2}\right]^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow\:\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}+\mathrm{4}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$${one}\:{root}\:{is}\:{x}=−\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{other}\:{root}\:=−\frac{\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}+\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:=\:−\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{2}}=\:−\mathrm{1} \\ $$$${y}=\mathrm{1} \\ $$$${pink}\:{segment}\:{length}=\mathrm{2} \\ $$
Commented by Tawa1 last updated on 04/Jun/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 04/Jun/19
Commented by mr W last updated on 04/Jun/19
$$\angle{DCA}=\angle{DCB} \\ $$$$\alpha=\frac{\mathrm{135}°}{\mathrm{2}}=\mathrm{67}.\mathrm{5}° \\ $$$$\beta=\mathrm{180}°−\mathrm{2}\alpha=\mathrm{45}° \\ $$$$\gamma=\mathrm{135}°−\beta=\mathrm{90}° \\ $$$${DE}=\sqrt{\mathrm{2}}{AE}=\sqrt{\mathrm{2}}×\sqrt{\mathrm{2}}=\mathrm{2} \\ $$$$ \\ $$$${or} \\ $$$${BCAD}\:{is}\:{lozenge},\:\Rightarrow{AD}//{BC} \\ $$$$\Rightarrow\beta=\mathrm{180}−\mathrm{135}=\mathrm{45}° \\ $$$$…… \\ $$
Commented by Tawa1 last updated on 04/Jun/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$