The graph y=kx+3 is tangent to the circle x^2 + y^2 = 7.
Find the possible values of k.
Find the possible values of k.
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x^2 + (kx + 3)^2 = 7
x^2 + k^2 * x^2 + 6kx + 9 = 7
x^2 * (1 + k^2) + 6kx + 2 = 0
In order for this line to be tangent (only intersect once), the discriminant of our quadratic must equal 0
b^2 - 4ac = 0
b = 6k
a = 1 + k^2
c = 2
(6k)^2 - 4 * (1 + k^2) * 2 = 0
36k^2 - 8 * (1 + k^2) = 0
36k^2 - 8 - 8k^2 = 0
28k^2 = 8
7k^2 = 2
k^2 = 2/7
k = +/- sqrt(2/7)
k = +/- (1/7) * sqrt(14)
x^2 + k^2 * x^2 + 6kx + 9 = 7
x^2 * (1 + k^2) + 6kx + 2 = 0
In order for this line to be tangent (only intersect once), the discriminant of our quadratic must equal 0
b^2 - 4ac = 0
b = 6k
a = 1 + k^2
c = 2
(6k)^2 - 4 * (1 + k^2) * 2 = 0
36k^2 - 8 * (1 + k^2) = 0
36k^2 - 8 - 8k^2 = 0
28k^2 = 8
7k^2 = 2
k^2 = 2/7
k = +/- sqrt(2/7)
k = +/- (1/7) * sqrt(14)