Example of Tangents Drawn from Point to the Circle

To find the equation of tangents 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 circle

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


Compare this circle with general equation of circle as {x^2} + {y^2} + 2gx + 2fy + c = 0
Here we have following values g = 1,\,\,\,f =   - 1,\,\,\,c = 1
Now centre of the circle (i) is \left( { - g, - f} \right) = \left( { - 1, - \left(  { - 1} \right)} \right) = \left( { - 1,1} \right)
Radius of 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, by 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 one equation of required tangents to the circle.
Putting the value of 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 second equation of required tangents to the circle.