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

Explanation

Given,`x/3+y/4=1`

We must have
=`((c/3+c/4-1)/sqrt(1/3^2+1/4^2))=c`

⇒`(12/5)(c/3​+c/4​−1)=±c`

`⇒c=6,1`