The equation of the circle which touches the axes of coordinates and the line `x/3+y/4=1` and whose center lies in the first quadrant is `x^2+y^2-2c x-2c y+c^2=0` , where `c` is (a) 1 (b) 2 (c) 3 (d) 6

Explanation

To solve the problem, we need to find the value of \( c \) for the equation of the circle that touches the coordinate axes and the line \( \frac{x}{3} + \frac{y}{4} = 1 \), with its center in the first quadrant.

### Step-by-Step Solution:

1. **Identify the center and radius of the circle:**
The equation of the circle is given as:
\[
x^2 + y^2 - 2cx - 2cy + c^2 = 0
...