(x-2)^2/4 + (y-k)^2/16 = 1
(x-1)^2 + (y+3)^2 = 1
We have an ellipse with center at (2,k) and radii 2 and 4
and a circle with center (1,-3) with radius 1
They are tangent internally at (0,-3) if k = -3
If they are tangent externally, then there will be two values at k = -3±c for some value of c.
So, regardless of just what c is, I get -9 as the sum of all possible values of k.
Find the sum of all possible values of the constant k such that the graph of the parametric equations x=2+4cos(s) and y=k-4sin(s) intersects the graph of the parametric equations x=1+cos(t) and y=-3+sin(t) at only one point.
3 answers
-9 is incorrect, sorry. The answer is also not 6 or -6
For those of you who want to know, it's -12