# 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

Now to solve this example we follow these steps.

Consider the given equation of a circle

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

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.:

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

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

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

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