A circle with a radius 4 touches the line 6x-8y=1. Find the locus of its center.

Let the centre of the circle be at
(x1,y1)).

Since the circle touches the line 6x-8y= 1, the
perpendicular distance of the centre (x1,y1) and line is 4. Also the perpendicular
distance of a point (h,k) from  a line ax+by+c = 0 is given by: d =
|ah+bh+c)|(a^2+b^2)^(1/2).

Therefore the  distance of
(x1,y1) from the line 6x-8y=1 or 6x-8y-1 = 0 is given by:

4
= |6x1-8y1-1|/{6^2+(-8)^2}^(1/2).

4 =
|6x1-8y1-1|/10.

6x1-8y-1 = 10*4, or 6x1-8y1-1 =
-10*4.

6x1-8y1-1-40 = 0 or 6x1-8y1-1+40 =
0.

Dropping the suffixes we get two lines
that are the locus of the center: 6x-8y-41 = 0 or 6x-8y+39 =
0