Points where a line intercepts a circle

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SUMMARY

The discussion focuses on finding the intersection points between the circle defined by the equation (x+2)² + y² = 1/2 and the vertical line x = -2. The user initially identified one intersection point at (-3/2, 1/2) but struggled to find the second point. A participant clarified that the solutions for x are x₁ = -3/2 and x₂ = -5/2, leading to the intersection points (-3/2, 1/2) and (-5/2, -1/2).

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circle: (x+2)^2+y^2=1/2
line: x+2
iv been able to find one point but can't find the other
work:
2(x+2)^2 =1/2
divide by 2 on both sides
(x+2)^2=1/4
square both sides
x+2=.5
subtract 2
x=-3/2
i used that to find the y but that only gives me one point please help
 
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thazel345 said:
circle: (x+2)^2+y^2=1/2
line: x+2
iv been able to find one point but can't find the other
work:
2(x+2)^2 =1/2
divide by 2 on both sides
(x+2)^2=1/4
square both sides
x+2=.5
subtract 2
x=-3/2
i used that to find the y but that only gives me one point please help

(Wave)

From $(x+2)^2=\frac{1}{4}$ we get that $x+2= \pm \frac{1}{2}$.
So $x_1=\frac{1}{2}-2=-\frac{3}{2}$ and $x_2=-\frac{1}{2}-2=-\frac{5}{2}$.
So we get the points $(x_1, x_1+2)=\left( -\frac{3}{2}, \frac{1}{2}\right)$ and $(x_2, x_2+2)=\left( -\frac{5}{2}, -\frac{1}{2}\right)$.
 

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