Let the radius of the circle be r. since, circle touches the lines x=0,y=0, i.e., x -axis and y -axis, then its centre is C≡(r,r)
Now, radius (PC)= Perpendicular distance from (C) to the line 4x+3y=12 i.e., ‌‌r=‌
|4r+3r−12|
√16+9
⇒‌‌7r−12=±5r So r=1 or 6 But r≠6 ∴ Required equation of circle is (x−1)2+(y−1)2=1 ⇒‌‌x2+1−2x+y2+1−2y=1 ⇒‌‌x2+y2−2x−2y+1=0