SOLUTION: For what value/s of k will the system of equations x^2 + y^2 = 4 and y=kx+4 have exactly two points?

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Question 1056901: For what value/s of k will the system of equations x^2 + y^2 = 4 and y=kx+4 have exactly two points?
Answer by ikleyn(52879) (Show Source):
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For what value/s of k will the system of equations x^2 + y^2 = 4 and y=kx+4 have exactly two points?
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= 4, (1) y = kx + 4. (2) Substitute the expression y = kx + 4 from (2) into equation (1). You will get = 4. Simplify = 4, = 0. In order for the last equation has two different real roots, the discriminant must be positive: d = b^2 - 4ac = (8k)^2 - 4*(1+k^2)*12 > 0, or 64k^2 - 48k^2 - 48 > 0, or 16k^2 > 48, or k^2 > 3, or k < OR k > . Answer. The solution set is (,) U (,).