Finding Intersection of Line and Circle: A Challenging Problem

[pokemeharder]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[courtrigrad]

set them equal to each other.

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

 

[pokemeharder]

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

 

[pokemeharder]

oh sry didnt see that

 

[pokemeharder]

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

 

[radou]

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

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.

 

[pokemeharder]

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

 

[radou]

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

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?

 

[pokemeharder]

well i graphed it and it showed no intersection

 

[pokemeharder]

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

 

[radou]

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

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.

 

[pokemeharder]

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

 

[Hurkyl]

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

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

 

[d_leet]

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

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.

 

[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

 

[d_leet]

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

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

 

[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)

 

[d_leet]

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)

You can't do that.

 

[pokemeharder]

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

 

[pokemeharder]

you must be an idiot if you can't get this. its quite damn simple...read your textbook

Im only 14...

 

[d_leet]

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

Was this directed at me? If so I know quite well what i is, I am saying that there should be NO i in your quadratic equation and that you cannot manipulate radicals as you did above.

 

[pokemeharder]

thats why it doesn't work out

its not quadratic anyways there's no x squared

 

[d_leet]

thats why it doesn't work out

its not quadratic anyways there's no x squared

It would be a quadratic if you had delt with the square root correctly? How do you undo a square root?

 

[pokemeharder]

oh yeah right...
didnt see that
how do u deal with the root properly then lol

 

[d_leet]

oh yeah right...
didnt see that
how do u deal with the root properly then lol

Well what happens if you square the square root of something.

 

[pokemeharder]

Well what happens if you square the square root of something.

XD right sry i feel stupid now
um thanks alot

 

[d_leet]

XD right sry i feel stupid now
um thanks alot

Your welcome.

 

[HallsofIvy]

I'm getting into this a bit late but I'm afraid Courtrigrads original suggestion of writing y= \sqrt{25-x^2} was misleading. Better is to substitute y= (1/4)(3x- 35) into x^2+ y^2= 25. Then you get a quadratic equation for x and can use the quadratic formula to show that it has no real solutions.