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Question Number 29365 by gyugfeet last updated on 08/Feb/18

the line x+y=2 cuts a circle x^2 +y^(2 ) =4 at two points. find the co−ordinates of points.

$${the}\:{line}\:{x}+{y}=\mathrm{2}\:{cuts}\:{a}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}\:} =\mathrm{4}\:{at}\:{two}\:{points}.\:{find}\:{the}\:{co}−{ordinates}\:{of}\:{points}. \\ $$

Answered by Rasheed.Sindhi last updated on 08/Feb/18

{ ((x+y=2)),((x^2 +y^(2 ) =4)) :}⇒ { ((y=2−x)),((x^2 +(2−x)^2 =4)) :} ⇒2x^2 −4x+4=4⇒x^2 −2x=0 ⇒x=0,2⇒y=2−0,2−2=2,0 (x,y)=(0,2) , (2,0) (0,2) and (2,0) are the coordinates of intersection points.

$$\begin{cases}{{x}+{y}=\mathrm{2}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}\:} =\mathrm{4}}\end{cases}\Rightarrow\begin{cases}{{y}=\mathrm{2}−{x}}\\{{x}^{\mathrm{2}} +\left(\mathrm{2}−{x}\right)^{\mathrm{2}} =\mathrm{4}}\end{cases} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}=\mathrm{4}\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0},\mathrm{2}\Rightarrow{y}=\mathrm{2}−\mathrm{0},\mathrm{2}−\mathrm{2}=\mathrm{2},\mathrm{0} \\ $$$$\left({x},{y}\right)=\left(\mathrm{0},\mathrm{2}\right)\:,\:\left(\mathrm{2},\mathrm{0}\right) \\ $$$$\left(\mathrm{0},\mathrm{2}\right)\:\mathrm{and}\:\left(\mathrm{2},\mathrm{0}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{coordinates}\:\mathrm{of} \\ $$$$\mathrm{intersection}\:\mathrm{points}. \\ $$

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