sub y = mx + b into x^2 + y^2 = r^2
expanding gave me
x^2 + m^2x^2 + 2bmx + b^2 - r^2 = 0
a^2(1+m^2) + 2bmx + b^2 - r^2 = 0
comparing this to Ax + By + C = 0
A = 1+m^2
B = 2bm
C = b^2 - r^2
we want this quadratic to have only one solution, since there is only one point of contact.
(That was the given hint)
So the discriminant of the quadratic formula must be zero, B^2 - 4AC = 0
(2bm)^2 - 4(1+m^2)(b^2 - r^2) = 0
4b^2m^2 - 4(1+m^2)(b^2 - r^2) = 0
b^2m^2 - (1+m^2)(b^2 - r^2) = 0
b^2m^2 - b^2 + r^2 - b^2m^2 + r^2m^2 = 0
- b^2 + r^2 + r^2m^2 = 0
r^2(1 + m^2) = b^2
as requested.
The tangent to a circle may be defined as the line that intersects the circle in a single point, called the point of tangency. If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show
r^2(1 + m^2) = b^2
Hint:
The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly one solution.
1 answer