Why we don't get two tangents in the point form of tangent from a given point to a circle? In my textbook I have read the point form of representation of tangent from a point $P(x_1,y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ which is given by 
$$xx_1 + yy_1 + g(x+x_1) + f(y+y_1)+c=0$$
I know that two tangents can be drawn from a point to a circle,but the above equation has only one equation of tangent.
Why the above equation holds for only one tangent when it should give two equations of tangents?
(I agree that there might be some inconsistencies in the equation.)
 A: The line that you mentioned is the line defined by the two points at which the tangent lines touch the circle. So, if you want to actually get those two points, compute the intersection between the line and the circle. Then, the tangent lines will be the lines passing through $(x_1,y_1)$ and each of the touching points.
A: I think what you have found in the text should be:-
If $P(x_1, y_1)$ is a point on the circle $C: x^2 + y^2 +2gx + 2fy + c = 0$, then the equation of the tangent touching circle C  at P is
$$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$$

If $P(x_0, y_0)$ is external to $C$, the equation(s) of tangents from P to C will not be a nice-looking simplified form. However, they can still be derived through the following steps.


*

*By midpoint formula, find M, the midpoint of OP. 

*By distance formula, find $R = \dfrac {OP}{2}$.

*Setup the equation of the new circle (centered at M and radius = R)

*Solve circle C and .circle M to get $H(x_1, y_1)$ and $K(x_2, y_2)$.

*Use two-point form to find the equation of $PH$ and $PK$, which are the required tangents.
