Given: Equation of circle is S:x2+y2−2x−4y−31=0 and the point P(5,2) Here, x1=5,y1=2 First let's find out S(x1,y1) ⇒S(x1,y1)=x12+y12−2x1−4y1−31=0 ⇒S(5,2)=25+4−10−8−31=−20<0 As we know that, if S(x1,y1)<0 for point P(x1,y1) then the point P lies inside the circle S. Hence, option A is the correct answer.