First find the slope/intercept equation of the line:Originally Posted bytigerdivision

x = -t + 600

y = 2t - 300

From the first equation t = 600 - x. So insert this into the y equation:

y = 2[600 - x] - 300

y = -2x + 900

The equation of the circle of radius 200 centered on the point (200, 100) is:

(x - 200)^2 + (y - 100)^2 = 200^2

We wish to find the intersection points of the line and the circle.

From the line equation we know that y = -2x + 900, so inserting this into the circle equation:

(x - 200)^2 + ([-2x + 900] - 100)^2 = 40000

x^2 - 400x + 40000 + (-2x + 800)^2 = 40000

x^2 - 400x + 4x^2 - 3200x + 640000 = 0

5x^2 - 3600x + 640000 = 0

x^2 - 720x + 128000 = 0

Whenever faced with large coefficients in a quadratic equation I just give up on trying to factor it and resort to the quadratic formula immediately. (I'm being a chicken about it, but it saves time. )

x = [-(-720) (+/-) sqrt{(-720)^2 - 4*1*128000}]/(2*1)

x = [720 (+/-) sqrt{6400}]/2

x = [720 (+/-) 80]/2

So x = 400 or x = 320

Thus

y = -2(400) + 900 = 100

or

y = -2(320) + 900 = 260

So the line intersects the circle at the points

(320, 260)

and

(400, 100)

-Dan