SOLUTION: a line y=c-3x is the tangent to a circle: x^2+y^2-4x-2y-5=0. What are the values of c?

Question 803198: a line y=c-3x is the tangent to a circle: x^2+y^2-4x-2y-5=0.
What are the values of c?
Answer by oscargut(2103) (Show Source): You can put this solution on YOUR website!
y=c-3x is the tangent to a circle: x^2+y^2-4x-2y-5=0
x^2+(c-3x)^2 -4x-2(c-3x)-5 =0
x^2+9x^2-6cx+c^2 -4x-2c+6x-5 =0
10x^2+(2-6c)x+c^2-2c-5 =0
has only one solution (tangent)if:
(2-6c)^2-4(10)(c^2-2c-5) = 0
-4c^2+56c+204 =0
c^2-14c-51 =0
c = (14+20)/2 or c = (14-20)/2
Answer: c =17 or c = -3
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