UPSC NDA Math Model Paper 4

Given: Equation of circle is S:x2+y2−6x−10y+26=0 and the point P(2,4)
Here, x1=2,y1=4
First let's find out S(x1,y1)
⇒S(x1,y1)=x12+y12−6x1 −10y1+26=0
⇒S(2,4)=4+16−12−40+26 =−6<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.