Length of the Tangent to a Circle

Let the tangent drawn from the point P\left( {{x_1},{y_1}} \right) meet the circle at the point T as shown in the given diagram. The equation is given by


length-tangent-to-circle

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

Consider the triangle PTC formed in this way is a right triangle, so according to the given diagram we have

{\left| {PT} \right|^2} + {\left| {TC} \right|^2} = {\left| {PC} \right|^2}\,\,\,\,{\text{ - - - }}\left( {{\text{ii}}} \right)

It is observed that \left| {TC} \right| is the radius of the circle, so {\left| {TC} \right|^2} = {g^2} + {f^2} - c.

We also have

{\left| {PC} \right|^2} = {\left( {{x_1} - \left( { - g} \right)} \right)^2} + {\left( {{y_1} - \left( { - f} \right)} \right)^2} = {\left( {{x_1} + g} \right)^2} + {\left( {{y_1} + f} \right)^2}

Putting all these values in (ii), we get

\begin{gathered} {\left| {PT} \right|^2} + {g^2} + {f^2} - c = {\left( {{x_1} + g} \right)^2} + {\left( {{y_1} + f} \right)^2} \\ \Rightarrow {\left| {PT} \right|^2} + {g^2} + {f^2} - c = {x_1}^2 + 2g{x_1} + {g^2} + {y_1}^2 + 2f{y_1} + {f^2} \\ \Rightarrow {\left| {PT} \right|^2} = {x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c \\ \Rightarrow \left| {PT} \right| = \sqrt {{x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c} \\ \end{gathered}

This gives the length of the tangent from the point P\left( {{x_1},{y_1}} \right) to the circle {x^2} + {y^2} + 2gx + 2fy + c = 0.

Similarly, we can show that the PS is also of the same length.

Example: Find the length of the tangent from \left( {12, - 9} \right) to the circle

3{x^2} + 3{y^2} - 7x + 22y + 9 = 0

Dividing the equation of the circle by 3, we get the standard form

{x^2} + {y^2} - \frac{7}{3}x + \frac{{22}}{3}y + 3 = 0

The required length of the tangent from \left( {12, - 9} \right) is

\sqrt {{{\left( {12} \right)}^2} + {{\left( { - 9} \right)}^2} - \frac{7}{3}\left( {12} \right) + \frac{{22}}{3}\left( { - 9} \right) + 3} = \sqrt {144 + 81 - 28 - 66 + 3} = \sqrt {134}