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?

Algebra ->  Circles -> 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?      Log On


   



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(52788) About Me  (Show Source):
You can put this solution on YOUR website!
For what value/s of k will the system of equations x^2 + y^2 = 4 and y=kx+4 have exactly two points?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

x%5E2+%2B+y%5E2 = 4,      (1)
y = kx + 4.       (2)

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

x%5E2+%2B+%28kx%2B4%29%5E2 = 4.

Simplify

x%5E2+%2B+k%5E2x%5E2+%2B+8kx+%2B+16 = 4,

%281%2Bk%5E2%29%2Ax%5E2+%2B+8kx+%2B+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%283%29  OR  k > sqrt%283%29.

Answer. The solution set is (-infinity,-sqrt%283%29) U (sqrt%283%29,infinity).