KEAM 2013 Math Paper

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