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

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

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