UPSC NDA Math Model Paper 5

Given: Equation of circle is S:x2+y2−8x−2y+12=0 and the point P (2, 2)
Here, x1=2,y1=2First let's find out S(x1,y1)
⇒S(x1,y1)=x12+y12−8x1 −2y1+12=0
⇒S(2,2)=4+4−16−4+12=0
As we know that, if S(x1,y1)=0 for point P(x1,y1) then the point P lies on the circle S.
Hence, option B is the correct answer.