# Homework Help: Circle - Line Intersection

1. Dec 3, 2006

### pokemeharder

1. The problem statement, all variables and given/known data

I can't find the intersection for Line y = 3/4x - 35/4
and Circle y^2 + x^2 = 25

2. Relevant equations

3. The attempt at a solution

2. Dec 3, 2006

set them equal to each other.

$$y^{2} = 25 - x^{2}$$

$$y =\pm \sqrt{25-x^{2}}$$

Last edited: Dec 3, 2006
3. Dec 3, 2006

### pokemeharder

how do i get rid of the square on y after i do that

4. Dec 3, 2006

### pokemeharder

oh sry didnt see that

5. Dec 3, 2006

### pokemeharder

ah can someone do this for me please, i'm stuck

6. Dec 3, 2006

Courtrigrad pretty much solved it for you already - all you have to do is set the equations equal, since the equation of the line is already given in the form y(x) = ...

Edit: draw a sketch first, this will pretty much solve your problem.

Last edited: Dec 3, 2006
7. Dec 3, 2006

### pokemeharder

i kno but i gor to 0 = -3/4x + 5ix + 35/4 and i don't really kno how to solve that

8. Dec 3, 2006

OK, let's slow down. Do you know how to make a sketch of the circle y^2 + x^2 = 25, and the line y = 3/4x - 35/4 ? The circle is centered at the origin, with radius 5, and you can sketch down the line easily by finding the points of intersection with the x and y axis (i.e. setting y = 0, and x = 0). What does that sketch tell you?

9. Dec 3, 2006

### pokemeharder

well i graphed it and it showed no intersection

10. Dec 3, 2006

### pokemeharder

now i just need to show that the circle and the line don't intersect algrebraecally or however you spell it.

11. Dec 3, 2006

Yes, after reading the posts above again, you'll end up with a quadratic equation which has no real solutions, which is what you need to show.

12. Dec 3, 2006

### pokemeharder

so wait this function 0 = -3/4x + 5ix + 35/4 has no real solutions because there is a complex number right?

13. Dec 3, 2006

### Hurkyl

Staff Emeritus
Incorrect -- sometimes complex equations can have real solutions. You have to show the solutions are not real. (say, by solving it)

Last edited: Dec 3, 2006
14. Dec 3, 2006

### d_leet

I'm wondering how you arrived at that equation, because I don't think you should end up with a quadratic equation with complex coefficients in this problem.

15. Dec 3, 2006

### pokemeharder

well if u set the y values to equal each other
3/4x + 35/4 = sqr root of (25 - x^2)

you get 0 = -3/4x + 5ix + 35/4

16. Dec 3, 2006

### d_leet

Ummm... I don't think thats what you get. I don't see where the 5ix comes from at all.

17. Dec 3, 2006

### pokemeharder

well u change it into sqr(25) x sqr(-x^2)
then u sqroot the 25 to 5 and the sqr root of -x^2 is sqr(-1) x sqr(x^2)
so its 5ix
i = sqr(-1)

18. Dec 3, 2006

### d_leet

You can't do that.

19. Dec 3, 2006

### pokemeharder

the square root of -1 is called i
its always called i

20. Dec 3, 2006

### pokemeharder

Im only 14......