TS EAMCET 20-July-2022 Shift 1 Solved Paper

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