Question

How are the points $(-1,1),(-2,2)$ and $(-2,3)$ situated with respect to the circle $x^2+y^2-4x–2y+3=0$

Answer 1

Given equation is $x^2+y^2+2x–4y-3=0$

Let $A=(-1,1),B=(-2,2)$ and $C=(-2,3)$

For point $A(-1,1),x^2+y^2+2x–4y-3=1+1-2-4-3=-7<0$

Hence point $A$ lies inside the circle

For point $B=(-2,2),x^2+y^2+2x–4y-3=4+4-4+8-3=9>0$

Hence point $B$ lies outside the circle

For point $C=(-2,3),x^2+y^2+2x–4y-3=4+9-4-12-3=-6<0$

Hence point $C$ lies inside the circle