UPSC NDA Math Model Paper 6

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.