Question

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

Answer 1

Given equation is $x^2+y^2–2x–4y+3=0$

Let $A=(0,1),B=(3,1)$ and $C=(1,3)$

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

Hence point $A$ lies on the circle

For point $B=(3,1),x^2+y^2–2x–4y+3=9+1–6–4+3=3>0$

Hence point $B$ lies outside the circle

For point $C=(1,3),x^2+y^2–2x–4y+3=1+9–2–12+3=-1<0$

Hence point $C$ lies inside the circle