Expert: Abe Mantell - 7/6/2009

Question
Let m be a real number and consider the line y=3+m(x-2) and the unit circle (x^2)+(y^2)=1 on the xy-plane. For what values m does the line intersect the circle exactly once? (The hint is that the answer should not be approximated)

Answer
We can solve this algebraically, although it does get messy.
Let's substitute y=3+m(x-2) into x^2 + y^2 =1.
Thus, we get: x^2 + (3+mx-2m)^2 = 1
==> x^2+9+6mx-12m+m^2*x^2-4m^2*x+4*m^2=1
==> (1+m^2)x^2+(6m-4m^2)x+4m^2-12m+8=0
Now using the quadratic formula, in order for there to be only one
solution, the discriminant (i.e. b^2-4ac) must be zero.  Hence,
(6m-4m^2)^2-4(1+m^2)(4m^2-12m+8)=0...which simplifies to:
-12m^2+48m-32=0 or 3m^2-12m+8=0, which yields:
m=2-(2/3)sqrt(3) and 2+(2/3)sqrt(3).

Those two slopes will give tangents to the circle.

Abe