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

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### 2 Answers

The circle with radius equal to 4 touches the line 6x - 8y = 1. This implies that the center of the circle lies at a distance of 4 from the line.

Taking a point (x,y) on the locus of the center, we use the relation for determining the distance d of a point (x1, y1) from the line ax+by +c = 0, which is:

d = |ax1+by1+c|/ sqrt (a^2+b^2)

4 = |6x -8y - 1|/sqrt ( 36 + 64)

=> 4 = |6x -8y - 1|/sqrt ( 100)

=> 4*10 = 6x - 8y - 1 and -40 = 6x - 8y - 1

=> 6x - 8y - 41 = 0 and

= > 6x - 8y + 39 = 0

**Therefore the locus of the center are the lines: **

**6x - 8y - 41 = 0 and 6x - 8y + 39 = 0**

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**