# Example of Tangents Drawn from a Point to the Circle

We will find the equation of tangent lines drawn from the point $\left( {1,5} \right)$ to the circle given by

${x^2} + {y^2} + 2x – 2y + 1 = 0$

Now to solve this example we follow these steps.

Consider the given equation of a circle

${x^2} + {y^2} + 2x – 2y + 1 = 0\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Compare this circle with the general equation of a circle as ${x^2} + {y^2} + 2gx + 2fy + c = 0$

Here we have the following values $g = 1,\,\,\,f = – 1,\,\,\,c = 1$

Now the center of the circle (i) is $\left( { – g, – f} \right) = \left( { – 1, – \left( { – 1} \right)} \right) = \left( { – 1,1} \right)$

The radius of the circle (i) is $r = \sqrt {{g^2} + {f^2} – c} = \sqrt {{{\left( { – 1} \right)}^2} + {{\left( 1 \right)}^2} – 1} = \sqrt 1 = 1$

Let $m$ be the slope of the tangent drawn from $\left( {1,5} \right)$ to the given circle (i), then its equation is

$\begin{gathered} y – 5 = m\left( {x – 1} \right) \\ \Rightarrow mx – y – m + 5 = 0\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered}$

Since the line (ii) is a tangent to the circle (i), so the distance of the centre of the circle should be equal to its radius, i.e.:

$\begin{gathered} \frac{{\left| {m\left( { – 1} \right) – \left( 1 \right) – m + 5} \right|}}{{\sqrt {{m^2} + {{\left( { – 1} \right)}^1}} }} = 1 \\ \Rightarrow \left| { – 2m + 4} \right| = \sqrt {{m^2} + 1} \\ \Rightarrow {\left( { – 2m + 4} \right)^2} = {m^2} + 1 \\ \Rightarrow 4{m^2} – 16m + 16 = {m^2} + 1 \\ \Rightarrow 3{m^2} – 16m + 15 = 0 \\ \end{gathered}$

Now solving this quadratic equation of variable $m$, using the quadratic formula we have

$\begin{gathered} \Rightarrow m = \frac{{ – \left( { – 16} \right) \pm \sqrt {{{\left( { – 16} \right)}^2} – 4\left( 3 \right)\left( {15} \right)} }}{{2\left( 3 \right)}} = \frac{{16 \pm \sqrt {256 – 180} }}{6} \\ \Rightarrow m = \frac{{18 \pm \sqrt {19} }}{3} \\ \end{gathered}$

Putting the value of one root, $m = \frac{{8 + \sqrt {19} }}{6}$, in equation (ii) we get

$\begin{gathered} \left( {\frac{{8 + \sqrt {19} }}{6}} \right)x – y – \left( {\frac{{8 + \sqrt {19} }}{6}} \right) + 5 = 0 \\ \Rightarrow \left( {8 + \sqrt {19} } \right)x – 3y – \left( {8 + \sqrt {19} } \right) + 15 = 0 \\ \Rightarrow \left( {8 + \sqrt {19} } \right)x – 3y – 8 – \sqrt {19} + 15 = 0 \\ \Rightarrow \left( {8 + \sqrt {19} } \right)x – 3y + 7 – \sqrt {19} = 0 \\ \end{gathered}$

This is the first equation of the required tangents to the circle.

Putting the value of the second root, $m = \frac{{8 – \sqrt {19} }}{6}$, in equation (ii) we get

$\begin{gathered} \left( {\frac{{8 – \sqrt {19} }}{6}} \right)x – y – \left( {\frac{{8 – \sqrt {19} }}{6}} \right) + 5 = 0 \\ \Rightarrow \left( {8 – \sqrt {19} } \right)x – 3y – \left( {8 – \sqrt {19} } \right) + 15 = 0 \\ \Rightarrow \left( {8 – \sqrt {19} } \right)x – 3y – 8 + \sqrt {19} + 15 = 0 \\ \Rightarrow \left( {8 – \sqrt {19} } \right)x – 3y + 7 + \sqrt {19} = 0 \\ \end{gathered}$

This is the second equation of the required tangents to the circle.