Position of a Point Relative to a Circle

For circle S = 0 and point P(x₁, y₁): S₁ > 0 ⟹ P outside; S₁ = 0 ⟹ P on the circle; S₁ < 0 ⟹ P inside.

Let the circle be $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$, with centre $C(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$.

For point $P(x_1, y_1)$, define:

$$S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$$

The distance from $C$ to $P$ is $d = \sqrt{(x_1+g)^2 + (y_1+f)^2}$.

Expanding $d^2$:

$$d^2 = x_1^2 + 2gx_1 + g^2 + y_1^2 + 2fy_1 + f^2 = S_1 + (g^2 + f^2 - c) = S_1 + r^2$$

Therefore $d^2 - r^2 = S_1$.

• $S_1 > 0 \Rightarrow d > r \Rightarrow P$ is outside the circle
• $S_1 = 0 \Rightarrow d = r \Rightarrow P$ is on the circle
• $S_1 < 0 \Rightarrow d < r \Rightarrow P$ is inside the circle

This is the standard test. It avoids computing $d$ explicitly — only the sign of $S_1$ matters.