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  • The point (1, 2) lies inside the circle x^2 + y^2 – 2x + 6y + 1 = 0.

 

The point (1, 2) lies inside the circle x2 + y2 – 2x + 6y + 1 = 0.

Answers (1)

False 

  Given circle is x2+y2-2x+6y+1=0 or (x-1)2+(y+3)2=32  

 Centre is C(1,-3) and radius is 3.  

 Distance of point P (1,2) from centre is 5.

 Thus CP>radius  

 So, point P lies outside the circle.

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