# 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

Now to solve this example we follow these steps
Consider the given equation of circle

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

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 $m$, by quadratic formula, we have

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

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

This is second equation of required tangents to the circle.