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 as {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.


point-position-wrt-circle

The given equation of circle is

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


The centre 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 centre must be greater than its radius, i.e. \left| {CP} \right| > r. Similarly, the point P lies on the circle, then its distance from the centre must be equal to the radius of circle i.e. \left| {CP} \right| = r and P lies inside the circle, then its distance from the centre must be less than its radius i.e. \left| {CP} \right| < r. Combining these three situations, we note that the point P lies outside, on or inside the circle according as

\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, above formula status that 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.

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