Centre lies on the line x+y=5. Let centre be (x1,5−x1)
Perpendicular distance from centre C to y−5=0 Perpendicular distance from centre to line x−2=0 |5−x1−2|=|x1−5| ⇒3−x1=±(x1−5) Now 3−x1=x1−5 ⇒x1=4 and 3−x1=−x1+5⇒3≠5,x1∈φ So, centre (4,1) Radius =1 ∴ Area of circle =πr2=π×l2=π sq. units