+31 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.
If someone could send help, it would be much appreciated! :)
+130474 The slope of the given line = -1
The slope of a line tangent to the circle is given by
-x/ y
So setting this = to the slope of the line
-x / y = -1
x = y
So
x^2 + x^2 = k
2x^2 = k
x^2 = k/2
x = sqrt(k/2)
So
x + y =3k
sqrt (k/2) + sqrt (k/2) =3k
2sqrt (k/2) = 3k square both sides
4k/2 = 9k^2
2k = 9k^2
k = (9/2)k^2
(9/2)k^2 - k = 0
k ((9/2)k -1) = 0
k = 0 reject
k = 2/9
3k = 6/9 = 2/3
