How do i get the second point of intersection again?

[Silhorn]

Hi,

I have a circle with center (0,0) of radius 4
It's equation should be:
X^2 + Y^2 = 16

I also have a straight line which has gradient of 1 (45deg) which goes through (4,0) and (0,-4)
It's equation should be:
Y = X - 4

I am trying to use calculations to confirm that the line intersects with the circle at points (4,0) and (0,-4) but can only get (0,4). I forget how to calculate the other coordinate (0,-4).

This is how I calculated my first intersect.

Put Y=X-4 value into circle:
X^2 + (X-4)^2 = 16

Simplify down: (Is this part correct?)
2X^2 - 16 = 16
2X^2 = 32
X^2=16
X=4

Now if i use 4 as value X and put back into circle equation then I get 0:
4^2 + Y^2 = 16
16 + Y^2 = 16
Y^2 = 16-16 = 0

So coordinate is (4,0).

Now how do i calculate so i get the other intercept, (-4,0)?

[dbasnett]

Have you seen this

http://mathworld.wolfram.com/Circle-...ersection.html

[jemidiah]

X^2 + (X-4)^2 = 16
=> X^2 + X^2 - 8X + 16 = 16
=> 2X^2 = 8X

If X=0, the equation is satisfied, so Y=X-4=-4 gives the intersection point (0, -4). If X is not 0, divide it off to get the equation
2X = 8
=> X=4
which gives rise to solution (4, 0).

So, you forgot the cross term and inverted the sign of the constant term when multiplying out the quadratic. These mistakes cancel each other out when it comes to finding the X=4 solution, but drop the X=0 solution.

[Silhorn]

Thanks for your replies and the link.

I haven't done this stuff in a long time so I forgot how to do some of it.