Question 1133939
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The lines of the form y=3x+k have slope 3.<br>
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.<br>
(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.<br>
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.<br>
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.<br>
{{{x^2+y^2-8 = 0}}}
{{{y = sqrt(8-x^2) = (8-x^2)^(1/2)}}}   [choose the half of the circle with y positive]
{{{dy/dx = (1/2)(8-x^2)^(-1/2)(-2x) = -x/(sqrt(8-x^2))}}}<br>
{{{-x/sqrt(8-x^2) = 3}}}
{{{-x/3 = sqrt(8-x^2)}}}
{{{x^2/9 = 8-x^2}}}
{{{(10/9)x^2 = 8}}}
{{{x^2 = 72/10 = 36/5}}}
{{{x = -6/sqrt(5)}}}<br>
Now solve for the corresponding y value.<br>
{{{(-6/sqrt(5))^2+y^2 = 8}}}
{{{36/5+y^2 = 8 = 40/5}}}
{{{y^2 = 4/5}}}
{{{y = 2/sqrt(5)}}}<br>
The point on the half of the circle where y is positive is (-6/sqrt(5),2/sqrt(5))<br>
Now find the equation of the line with slope 3 through that point.<br>
{{{2/sqrt(5) = 3(-6/sqrt(5))+k}}}
{{{k = 20/sqrt(5) = 4*sqrt(5)}}}<br>
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....<br>
ANSWER: The line y=3x+k does not intersect the curve x^2+y^2-8=0 for value of k for which {{{abs(k)>4*sqrt(5)}}}