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.