4.0 Position of a point with respect to a circle

Let us assume the equation of circle in general form ${x^2} + {y^2} + 2gx + 2fy + c = 0$.

1. If point $P$ lies inside the circle, then ${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c < 0$.

2. If point $P$ lieson the circle, then ${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c = 0$.

3. If point $P$ lies outside the circle, then ${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c > 0$.

$S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0$ and ${S_1} \equiv {x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c$

Let $P({x_1},{y_1})$ be the given point and $C$ be the centre of the circle, $C \equiv \left( { - g, - f} \right)$

Distance between centre $C$ and point $P$ using distance formulae is $$CP = \sqrt {{{\left( {{x_1} + g} \right)}^2} + {{\left( {{y_1} + f} \right)}^2}} = \sqrt {{x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c} $$

Let $r$ be the radius of the circle, then $r = \sqrt {{g^2} + {f^2} - c} $

If point $P$ lies inside the circle, $CP<r$ i.e., squaring both sides we get $${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c < {g^2} + {f^2} - c$$

If point $P$ lies on the circle, $CP=r$ i.e., squaring both sides we get $${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c = {g^2} + {f^2} - c$$

If point $P$ lies outside the circle, $CP>r$ i.e., squaring both sides we get $${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c > {g^2} + {f^2} - c$$