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?
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{{{x^2 + y^2}}} = 4,      (1)
y = kx + 4.       (2)

Substitute the expression y = kx + 4 from (2) into equation (1). You will get

{{{x^2 + (kx+4)^2}}} = 4.

Simplify

{{{x^2 + k^2x^2 + 8kx + 16}}} = 4,

{{{(1+k^2)*x^2 + 8kx + 12}}} = 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 < {{{-sqrt(3)}}}  OR  k > {{{sqrt(3)}}}.

<U>Answer</U>. The solution set is ({{{-infinity}}},{{{-sqrt(3)}}}) U ({{{sqrt(3)}}},{{{infinity}}}).
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