TS EAMCET 19-July-2022 Shift 2 Solved Paper

Given, equation of circle is
x2+y2−2x−2y+1=0
Centre of circle =(1,1) and radius =1 unit
The point B lies on the circle, the coordinate of B is (1+cos‌θ,1+sin‌θ).
Given, coordinate of A is (0,−2)
(AB)2=(1+cos‌θ)2+(1+sin‌θ+2)2
‌=1+cos2θ+2‌cos‌θ+9+sin‌2θ+6sin‌θ
‌=11+2‌cos‌θ+6sin‌θ
As we know that
−√a2+b2≤a‌cos‌θ+bsin‌θ≤√a2+b2
So, −√4+36≤2‌cos‌θ+6sin‌θ≤√4+36
−2√10<2‌cos‌θ+6sin‌θ≤2√10
Maximum value of (AB)2 is 11+2√10.