+1167 Let k be a positive real number. The line x + y = 3k and the circle x^2 + y^2 = k - 4x - 6y are drawn. Find $k$ so that the line is tangent to the circle.
+129771 y = -x + 3k ......the slope of this line = -1
x^2 + 4x + y^2 + 6y = k complete the square on x,y
x^2 + 4x + 4 + y^2 + 6y + 9 = k + 4 + 9
(x + 2)^2 + (y + 3)^2 = 13 + k
The center of this circle is (-2, -3)
A line with a slope = 1 through this point will intersect the tangent line
y = (x +2) - 3
y = x -1
So
x -1 = -x + 3k
2x = 3k+1
x = (3k + 1) / 2
y = (3k+ 1) / 2 - 1
y = (3k + 1 - 2) / 2
y = (3k - 1) / 2
Putting these values into the equation of the circle to find k
[ (3k + 1) / 2 + 2]^2 +[ (3k -1)/2 + 3[^2 = 13 + k
Simplifying this we get
9k^2 + 28k - 1 = 0
Use the Quad Formula
We have two values for k
k ≈ .035313
and
k ≈ -3.1464
