UPSC NDA Math Model Paper 9

Given: Equation of circle is S:x2+y2−4x+6y+4=0 and the point P (- 1, - 3)
Here, x1=−1,y1=−3
First let's find out S(x1,y1)
⇒S(x1,y1)=x12+y12−4x1 +6y1+4=0
⇒S(−1,−3)=1+9+4−18+4=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.