For what values of b will the line y=2x+b be tangent to the circle x^2+y^2=9?

Answer 1

clearly the line must intersect the circle in only one point.

x^2 + (2x+b)^2 = 9
5x^2 + 4bx + b^2-9 = 0
The discriminant of this quadratic is

(4b)^2 - 4(5)(b^2-9) = -4b^2+180
For the quadratic to have a single solution, the discriminant must be zero, so
b^2 = 45
b = ±3√5
That means y=2x±3√5

See the graphs at

http://www.wolframalpha.com/input/?i=plot+x^2%2By^2%3D9%2C+y%3D2x%2B3%E2%88%9A5%2C+y%3D2x-3%E2%88%9A5

Or, you could note that for the circle,

y' = -x/y
So we must have

-x/y = 2
-x = 2y

4y^2+y^2=9
y^2 = 9/5
y = ±3/√5
x^2+9/5 = 9
x^2 = ∓6/√5

y = 2x+b
-3/√5 = 12/√5 + b
b = ±3√5