AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

S=x2+y2−2x−4y+3=0
And,
p‌‌=(−1)2+(1)2−2(−1)−4(1)+3
‌‌=3
And
t‌‌=√p
‌‌=√3
Now the circle with centre (p,t2) is (3,3)
(x−3)2+(y−3)2=r2
The circle passes through origin so,
(0−3)2+(0−3)2‌‌=r2
r2‌‌=9+9
‌‌=18
The circle S′ will be,
(x−3)2+(y−3)2=18
Now, the point (2,3) w.r.t circle (x−3)2+(y−3)2=18 is
(2−3)2+(3−3)2−18‌‌=1−18
‌‌=−17<0
So, the point (2,3) lies inside the circle S′=0