# 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.)

• Is it possible that the point (which is $(x_1,y_1$), I presume) is supposed to be on the circle? – Arthur Dec 24 '18 at 11:44
• $(x_1,y_1)$ are arbitary external point from which tangent is drawn(it can be anywhere except inside the circle). – pranjal verma Dec 24 '18 at 11:46
• This is the equation of the polar of $P$, which is tangent to the circle iff $P$ is on the circle. – amd Dec 25 '18 at 2:26

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.

• Actually, the slope of the line if $-\frac{y_1+f}{x_1+g}$. And what I wrote was that that line passes through both tangent points. Why should the slope be quadratic then? – José Carlos Santos Dec 24 '18 at 12:04
• Yes, that is what I mean. If, for instance, $f=g=0$, $c=-1$, and $(x_1,y_1)=(2,0)$, then the circle is the unit circle and the touching points are $\left(\frac12,\pm\frac{\sqrt3}2\right)$. Then the line that you get is the line $x=\frac12$, which passes through both points. – José Carlos Santos Dec 24 '18 at 12:19
• Did you actually read what I wrote? I wrote that the line that you mentioned is the line defined by the two tangent points. I did not claim that it is tangent to the circle. – José Carlos Santos Dec 24 '18 at 13:02
• @pranjalverma If my answer was useful, perhaps that you could mark it as the accepted one. – José Carlos Santos Dec 25 '18 at 10:23
• Sure. I missed that. Done. – José Carlos Santos Dec 25 '18 at 16:05

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.

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

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

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

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

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