+33 Let k be a positive real number. The line \($x + y = 3k$\) and the circle \($x^2 + y^2 = k$\) are drawn. Find k so that the line is tangent to the circle.
+1286 For the line and circle to be tangent, there must be exactly one point of intersection. If we substitute the equation of the line, y = 3k - x, into the equation of the circle, we can see if this is the case.
Substituting y: x^2 + (3k - x)^2 = k
Expanding the square: x^2 + 9k^2 - 6kx + x^2 = k Combining terms: 2x^2 - 6kx + 9k^2 - k = 0 Factoring: (x - 3k)(2x - 1) = 0
This equation has two solutions: x = 3k and x = 1/2. For there to be only one point of intersection, one of these solutions must be extraneous (not a valid point of intersection).
If x = 3k, then substituting back into the equation of the line, we get: 3k + y = 3k y = 0
This point, (3k, 0), lies on the circle for any value of k. Therefore, x = 3k is a valid point of intersection.
The other solution, x = 1/2, needs to be extraneous. For this to be true, the discriminant of the quadratic equation must be zero. The discriminant is:
b^2 - 4ac = (-6k)^2 - 4 * 2 * (9k^2 - k) = 36k^2 - 72k^2 + 4k = -36k^2 + 4k
Setting this equal to zero and solving for k: -36k^2 + 4k = 0 k(-36k + 4) = 0
k = 0 or k = 1/9
Since k is a positive real number, we reject k = 0. Therefore, k = 1/9 is the only value that makes the discriminant zero, resulting in one point of intersection between the line and the circle.
So, k = 1/9.
