# Position of a Point Relative to a Circle

The point $P\left( {{x_1},{y_1}} \right)$ lies outside, on or inside the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ according to ${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c > 0$, ${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c = 0$ and ${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c < 0$ respectively. The given equation of circle is

${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

The center of the circle (i) is $C\left( { – g, – f} \right)$ and its radius is $r = \sqrt {{g^2} + {f^2} – c} \,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

In the given diagram, three different positions of the point $P\left( {{x_1},{y_1}} \right)$ are shown. It is clear from the diagram that if the point $P$ lies outside the circle, then its distance from the center must be greater than its radius, i.e. $\left| {CP} \right| > r$. Similarly, the point $P$ lies on the circle, so its distance from the center must be equal to the radius of the circle i.e. $\left| {CP} \right| = r$ and $P$ lies inside the circle, so its distance from the center must be less than its radius i.e. $\left| {CP} \right| < r$. Combining these three situations, we note that point $P$ lies outside, on or inside the circle according to

$\left| {CP} \right|\frac{ > }{ < }r\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)$

Now

$\left| {CP} \right| = \sqrt {{{\left( {{x_1} – \left( { – g} \right)} \right)}^2} + {{\left( {y – \left( { – f} \right)} \right)}^2}} = \sqrt {{{\left( {{x_1} + g} \right)}^2} + {{\left( {{y_1} + f} \right)}^2}} \,\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right)$

Using the values of $r$ and $\left| {CP} \right|$ from (ii) from (iv) in (iii), we have

$\begin{gathered} \sqrt {{{\left( {{x_1} + g} \right)}^2} + {{\left( {{y_1} + f} \right)}^2}} \frac{ > }{ < }\sqrt {{g^2} + {f^2} – c} \\ \Rightarrow {\left( {{x_1} + g} \right)^2} + {\left( {{y_1} + f} \right)^2}\frac{ > }{ < }{g^2} + {f^2} – c \\ \Rightarrow {x_1}^2 + 2g{x_1} + {g^2} + {y_1}^2 + 2f{y_1} + {f^2}\frac{ > }{ < }{g^2} + {f^2} – c \\ \Rightarrow {x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c\frac{ > }{ < }0 \\ \end{gathered}$

NOTE: In fact, using the formula above to check the position of a point $P\left( {{x_1},{y_1}} \right)$ relative to a circle, we put the coordinates of the point in the equation of the circle.
(i) If the result is positive, then the point lies outside the circle.
(ii) If the result is zero, then the point lies on the circle.
(iii) If the result is negative, then the point lies inside the circle.