For what values of k is the line y=k tangent to the graph (x-4)^2 + y^2 = 16
Solution
The line y = k is tangent to the circle (x-4)^2 + y^2 = 16 if the center of the circle lies at a
distance from the line equal to the radius of the circle.
(x-4)^2 + y^2 = 16
=> (x - 4)^2 + (y - 0)^2 = 4^2
the center of the circle is (4, 0) and the radius is 4.
The distance of a point (x1 , y1) from a line ax + by + c = 0 is given by |a*x1 + b*y1 + c|/sqrt
(a^2 + b^2)
Using the values given:
4 = |4*0 + 0*1 - k|/sqrt (1)
=> |-k| = 4
=> k = 4 and k = -4
The lines y - 4 = 0 and y + 4 = 0 are tangents to the circle.