MHB Points where a line intercepts a circle

Click For Summary
The discussion focuses on finding the intersection points between the line x + 2 and the circle defined by (x + 2)² + y² = 1/2. One point, (-3/2, 1/2), was identified, but the second point remained elusive. A participant clarified that from the equation (x + 2)² = 1/4, two solutions for x can be derived: x = -3/2 and x = -5/2. Consequently, the second intersection point is determined to be (-5/2, -1/2). The thread concludes with both intersection points established.
thazel345
Messages
1
Reaction score
0
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
 
Mathematics news on Phys.org
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 $(x_2, x_2+2)=\left( -\frac{5}{2}, -\frac{1}{2}\right)$.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 59 ·
2
Replies
59
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
6
Views
2K
Replies
6
Views
1K
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K