Question 1133939: Find the range of values of k, such that the straight line does not intersect the curve . Found 2 solutions by greenestamps, JaydenX:Answer by greenestamps(13198) (Show Source):
The curve x^2+y^2-8=0 is a circle with center at the origin and radius 2*sqrt(2); there will be two points on the circle where the slope of the curve -- and therefore the slope of a tangent to the curve -- is 3.
(1) Find the points where the slope of the curve is 3.
(2) Find the equations of the lines tangent to the curve at those points. The values of k for which the line y=3x+k DOES intersect the circle are the values between the y-intercepts of those two lines.
Note: Because the origin is the center of the circle, the two solutions will be symmetrical; we only need to do the complete solution process for one of them.
If we choose the half of the circle with y positive, a line with slope 3 will have a positive y-intercept; and the x value will be negative.
[choose the half of the circle with y positive]
Now solve for the corresponding y value.
The point on the half of the circle where y is positive is (-6/sqrt(5),2/sqrt(5))
Now find the equation of the line with slope 3 through that point.
By symmetry, the two values of k for which the line with 3x+k just touches the circle are 4*sqrt(5) and -4*sqrt(5). So....
ANSWER: The line y=3x+k does not intersect the curve x^2+y^2-8=0 for value of k for which
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Using the formula as it doesn’t intersect the curve.
Separate into possible cases, and
Hence, the range of values of k, ,