SOLUTION: find the range of values of k for which kx+y=3 meets x�+y�=5 in two distinct points

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Question 1051725: find the range of values of k for which kx+y=3 meets x�+y�=5 in two distinct points
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website!
All the lines go through the point (0,3).
The line tangent to the circle and going through the point (0,3) would make a right triangle together with the radius of the circle with the hypotenuse being from the origin to the point (0,3).

So then build a circle centered at (0,3) with a radius of 2.
Then look for the two intersection points of that circle with the original circle.
Those two points are the two tangents points that the lines would make with the original circle and will provide the limits for .
The equation of the new circle would be,

So using the original equation,

Substitute this into the new equation,

So them.

So now you have two points on each line, you can solve for .

with the other solution for k being the reciprocal,

So the tangent points are the limit for one distinct point of contact so for two distinct points don't include these points so,

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Answer by ikleyn(52778) (Show Source):
You can put this solution on YOUR website!
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find the range of values of k for which kx+y=3 meets x�+y�=5 in two distinct points
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kx + y = 3 (1) x� + y� = 5 (2) Express y = 3-kx from (1) and substitute it into (2). You will get = 5. Simplify: = 5, = 0, = 0. Discriminant d = = = = = . The condition for "kx+y=3 meets x�+y�=5 in two distinct points" is this inequality d > 0, or > 0, or, which is the same (cancel the factor 4) > 0, or |k| > = . Answer. kx+y=3 meets x�+y�=5 in two distinct points if and only if k < OR k > .