# Thread: Find the point(s) of intersection.

1. ## Find the point(s) of intersection.

Algebra II.

Find the point(s) of intersection. List all of the solutions.

The line:
x -2y = 0

&

The circle:
(x-2)^2 + (y+2)^2 = 36

For the circle I think the radius is 6 and the center is (2,-2). From there I know that x = 2y but I get lost when substituting it in.

2. $(x-2)^2 + (y+2)^2 = 36$

$x = 2y$

Substitute in 2y everytime x appears in the circle equation:

$(2y-2)^2 + (y+2)^2 = 36$

$[4y^2 - 8y + 4] + [y^2 + 4y + 4] = 36$

$5y^2 - 4y - 28 = 0$

Now use the quadratic formula to solve for y. You should get 2 answers. Substitute in y back into either equation to get the x co-ordinate and these are your points of intersection.