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At-what-value-of-a-does-the-system-of-equations-have-4-solutions-x-2-y-2-0-x-a-2-y-2-1-




Question Number 209520 by Ismoiljon_008 last updated on 12/Jul/24
     At what value of  a does the system     of equations have 4 solutions ?     { x^2 −y^2 =0     {(x−a)^2 +y^2 =1
$$ \\ $$$$\:\:\:{At}\:{what}\:{value}\:{of}\:\:{a}\:{does}\:{the}\:{system} \\ $$$$\:\:\:{of}\:{equations}\:{have}\:\mathrm{4}\:{solutions}\:? \\ $$$$\:\:\:\left\{\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0}\right. \\ $$$$\:\:\:\left\{\left({x}−{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\right. \\ $$$$ \\ $$
Commented by Ismoiljon_008 last updated on 12/Jul/24
   help please
$$\:\:\:{help}\:{please} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Jul/24
 { ((x^2 −y^2 =0⇒x^2 =y^2 )),(((x−a)^2 +y^2 =1)) :}   x^2 −2ax+a^2 +y^2 =1  x^2 −2ax+a^2 +x^2 =1  2x^2 −2ax+a^2 −1=0  x=((2a±(√(4a^2 −4(2)(a^2 −1))))/4)  x=((2a±2(√(2−a^2 )))/4)  x=((a±(√(2−a^2 )))/2)⇒2−a^2 ≠0⇒a≠±(√2)  For all other values of a  x has two values  y^2 =x^2 ⇒y=±∣x∣  (((a+(√(2−a^2 )))/2) ,±((a+(√(2−a^2 )))/2))  (((a−(√(2−a^2 )))/2) ,±((a−(√(2−a^2 )))/2))  a∈ R−{±(√2) }
$$\begin{cases}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0}\Rightarrow{x}^{\mathrm{2}} ={y}^{\mathrm{2}} }\\{\left({x}−{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}}\end{cases}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{ax}+{a}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{ax}+{a}^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{ax}+{a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}{a}\pm\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)}}{\mathrm{4}} \\ $$$${x}=\frac{\mathrm{2}{a}\pm\mathrm{2}\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{4}} \\ $$$${x}=\frac{{a}\pm\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\Rightarrow\mathrm{2}−{a}^{\mathrm{2}} \neq\mathrm{0}\Rightarrow{a}\neq\pm\sqrt{\mathrm{2}} \\ $$$${For}\:{all}\:{other}\:{values}\:{of}\:{a} \\ $$$${x}\:{has}\:{two}\:{values} \\ $$$${y}^{\mathrm{2}} ={x}^{\mathrm{2}} \Rightarrow{y}=\pm\mid{x}\mid \\ $$$$\left(\frac{{a}+\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\:,\pm\frac{{a}+\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\right) \\ $$$$\left(\frac{{a}−\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\:,\pm\frac{{a}−\sqrt{\mathrm{2}−{a}^{\mathrm{2}} }}{\mathrm{2}}\right) \\ $$$${a}\in\:\mathbb{R}−\left\{\pm\sqrt{\mathrm{2}}\:\right\} \\ $$
Commented by Ismoiljon_008 last updated on 12/Jul/24
   thank you very much
$$\:\:\:{thank}\:{you}\:{very}\:{much}\: \\ $$

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