# Math Help - Intersection of line with circle

1. ## Intersection of line with circle

Hi
I have a circle x^2 + y^2=2, a point (1,1) on the circle, a point on the x-axis (t,0) where t is an arbitrary parameter and a straight line, L, passing through both points.
How would I calculate the coordinates of the second point of intersection of the line L with the circle?

Thanks

2. Originally Posted by bigdoggy
Hi
I have a circle x^2 + y^2=2, a point (1,1) on the circle, a point on the x-axis (t,0) where t is an arbitrary parameter and a straight line, L, passing through both points.
How would I calculate the coordinates of the second point of intersection of the line L with the circle?
The line passes through (1,1) and (t,0).
The center of the circle is (Xr,Yr)
The circle intersects the Line at (1,1) and (Xi,Yi)
The radius of the circle is 2.
What you are seeking is the coordinates of the intersection
of the line and circle at the 2nd point?
Is that what you mean?

3. Originally Posted by bigdoggy
Hi
I have a circle x^2 + y^2=2, a point (1,1) on the circle, a point on the x-axis (t,0) where t is an arbitrary parameter and a straight line, L, passing through both points.
How would I calculate the coordinates of the second point of intersection of the line L with the circle?

Thanks
1. The line L has the equation:

$\dfrac{y-1}{x-1} = \dfrac{0-1}{t-1}~\implies~y=\dfrac1{1-t} \cdot x + \dfrac t{t-1}$

2. Plug in the term of y into the equation of the circle and solve the equation for x. Keep in mind that one solution must be x = 1:

$x^2+\left(\dfrac1{1-t} \cdot x + \dfrac t{t-1} \right)^2=2$

3. After some premier league transformations you'll get:

$x=-\dfrac{t^2-4t+2}{t^2-2t+2}~\vee~x=1$

4. To calculate the y-coordinate of the second point plug in the x-value into the equation of L. You should get: $y = \dfrac{t^2-2}{t^2-2t+2}$

4. Thanks for your help, but is there a way to find the solution using vector methods?