Question 1209461
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x^2 + y^2 = 4
x^2 + (mx+c)^2 = 4 .................  plug in y = mx+c
x^2 + m^2x^2+2mcx+c^2 = 4
(1+m^2)x^2 + 2mcx + c^2-4 = 0


We have the quadratic function
f(x) = (1+m^2)x^2 + 2mcx + (c^2-4)
where,<ul><li>x^2 coefficient = (1+m^2)</li><li>x coefficient = 2mc</li><li>constant = (c^2-4)</li></ul>If the discriminant is 0, then the quadratic has exactly one root. 
This generates exactly one point of intersection between the circle and the tangent line.


d = discriminant
d = (x coefficient)^2 - 4*(x^2 coefficient)*(constant)
d = (2mc)^2 - 4*(1+m^2)(c^2-4)
(2mc)^2 - 4*(1+m^2)(c^2-4) = 0
4m^2c^2 - 4*(1+m^2)(c^2-4) = 0
4( m^2c^2 - (1+m^2)(c^2-4) ) = 0
m^2c^2 - (1+m^2)(c^2-4) = 0/4
m^2c^2 - (c^2-4) - m^2(c^2-4) = 0
m^2c^2 - (c^2-4) - m^2c^2+4m^2 = 0
-(c^2-4)+4m^2 = 0
4m^2 = c^2-4
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