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