the two graphs are
x^2 + y^2 = 1
(x-2)^2 + (y-3)^2 = 16
x^2-4x+4 + y^2-6y+9 = 16
but, x^2+y^2 = 1, so that means
-4x+4 + 1 -6y+9 = 16
4x+6y = -2
so the slope of the line is -2/3
No need to actually find the points.
The graph of the parametric equations
x=cos(t)
y=sin(t)
meets the graph of the parametric equations
x=2+4cos(s)
y=3+4sin(s)
at two points. Find the slope of the line between these two points.
Please help!!!! I need this done soon and I am getting nowhere with this problem
1 answer