For what values of b will the line y=2x+b be tangent to the circle x^2+y^2=9?
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
To find the values of b for which the line y = 2x + b is tangent to the circle x^2 + y^2 = 9, we need to determine the points of tangency between the line and the circle.
The equation of the line y = 2x + b can be rewritten as:
2x - y + b = 0
We can substitute the value of y from this equation into the equation of the circle to find the intersection points. Substituting y = 2x + b into x^2 + y^2 = 9, we get:
x^2 + (2x + b)^2 = 9
Expanding and simplifying the above equation yields:
x^2 + 4x^2 + 4bx + 4b^2 = 9
Combining like terms, we get:
5x^2 + 4bx + 4b^2 - 9 = 0
For the line to be tangent to the circle, this quadratic equation should have only one solution. In mathematical terms, this means the discriminant (b^2 - 4ac) of the quadratic equation should be equal to zero.
In our case, the coefficients of the quadratic equation are: a = 5, b = 4b, and c = 4b^2 - 9. So the discriminant becomes:
(b^2)^2 - 4(5)(4b^2 - 9) = 0
Simplifying further:
b^4 - 80b^2 + 180 = 0
At this point, we have a quartic equation. Solving a quartic equation can be quite complex, so it is better to use an alternative approach.
Let's observe that if a quadratic equation has a double root, then the discriminant is zero, which means b^4 - 80b^2 + 180 = 0 implies that there is at least one double root. This means we need to find the values of b for which b^4 - 80b^2 + 180 = 0.
Fortunately, this quartic equation can be factored as:
(b^2 - 20)(b^2 - 9) = 0
Setting each factor equal to zero, we get:
b^2 - 20 = 0 --> b^2 = 20 --> b = ±√20
b^2 - 9 = 0 --> b^2 = 9 --> b = ±3
So we have four possible values for b: √20, -√20, 3, -3.
Therefore, the line y = 2x + b will be tangent to the circle x^2 + y^2 = 9 for b = √20, -√20, 3, or -3.