Given a circle $C(x,y) \equiv x^2 + y^2 + 2gx+2fy+c=0$ and a point $P = (x_1,y_1)$ outside the circle, the equation of the pair of straight lines that are tangent to the circle and pass through $P$ is given by $$C(x_1,y_1)C(x,y) = T(x,y)^2,$$

where $T(x,y)$ is the equation of the chord of contact of the tangents drawn from point $P$.

I do not know a derivation using Plucker's $\mu$. Please help, this question is driving me nuts. I am unable to understand the significance of squaring the equation of a line.

  • $\begingroup$ Your equation does not make sense to me. C(x1,y1) is a constant, but different locations of P have completely different coefficients for polar and tangents. $\endgroup$ – coproc Oct 18 '15 at 19:35
  • $\begingroup$ I am talking about the equation of pair of tangents to circle through a point P. See iit-jee-maths.blogspot.in/2008/12/… $\endgroup$ – Isomorphism Oct 19 '15 at 4:33

To find the equation of the pair of tangents, we have to find a second degree curve which passes through the intersection of $C=0$ and $T=0$ (say $A$ and $B$) and is tangent to $C$.

For this consider the curve $S(x,y)\equiv C(x,y)+\lambda T^2(x,y)$.

This curve will obviously always pass through $A$ and $B$ because at those points both $C=0$ and $T=0$ hold true. Now the reason we have squared the equation of the line is because we want the required curve to be tangent to $C$ at both $A$ and $B$. Have a look at the derivative of $S$: $$\frac{d}{dx}\left(S(x,y)\right)=\frac{d}{dx}\left(C(x,y)\right)+2\lambda T(x,y)\frac{d}{dx}\left(T(x,y)\right)$$ You can see that at the points $A$ and $B$ the derivative of $S$ is equal to the derivative of $C$ and hence $S$ and $C$ must be 'touching' at these points. Had we not squared the line's equation, this would not have been possible due to the absence of the extra $T(x,y)$.

After understanding the above, all you have to do is plug in the co-ordinates of $P$ in the curve $S$ since it passes through that point, by doing which we get:- $C(x_1,y_1)+\lambda T^2(x_1,y_1)=0$

You can check that for any point $P$, $C(x_1,y_1)=T(x_1,y_1)$ and this will result in $\lambda=-\dfrac 1{C(x_1,y_1)}$. Plugging the value of $\lambda$ completes the derivation.

EDIT: A couple of clarificatons :

For the purpose of this question the assumption is that we're using "standard" forms of the equations for the circle and the chord of contact, ie,

$$C\equiv x^2+y^2+2gx+2fy+c$$ and $$T\equiv xx_1+yy_1+g(x+x_1)+f(y+y_1)+c $$

Keeping that in mind, it's fairly obvious that $C(x_1,y_1)=T(x_1,y_1)$

Additionally, you can easily check that the discriminant for the resultant conic $CC_1 = T^2$, is zero if $P$ is the origin (it is a tedious task to show this for a general point, but it can be done), so I'm fairly confident that it is always a pair of straight lines.

  • $\begingroup$ After writing this up, I am wondering why the curve $S$ must be a pair of lines. It could be a hyperbola for all we know. $\endgroup$ – G-man Oct 20 '15 at 6:10
  • $\begingroup$ I did not follow the derivative-tangent argument. Why should the equality of the curve equation's derivative imply tangency? $\endgroup$ – Isomorphism Oct 20 '15 at 9:17
  • $\begingroup$ @Isomorphism you can check that at the interection points of the curves the values of $\frac{dy}{dx}$ for both the curves are also equal. Hence the tangents at the points coincide, and the curves touch. $\endgroup$ – G-man Oct 20 '15 at 15:38
  • $\begingroup$ @G-man How did you conclude that for any point $P$, $C(x_1,y_1)$=$T(x_1,y_1)$? What I can see is that $T(x_1,y_1)=0$ as $T$ passes through it whereas $C(x_1.y_1) \neq 0$. What am I missing? $\endgroup$ – MathGeek Aug 28 '16 at 10:46
  • $\begingroup$ How did you conclude that the given second degree equation represents a pair of straight lines only? $\endgroup$ – Sharv Laad Oct 10 '18 at 13:25

A more general proof:

Let Q and R be the points at which lines through $P=(x_1,y_1)$ touch a non degenerate conic $S(x,y) \equiv Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$. In other words, lines PR and PQ are the tangents to this conic at points Q and R, and RQ is the chord of contact.

Let $PR(x,y)=0$, $PQ(x,y)=0$, $RQ(x,y)=0$ be the equation of these lines.

As RQ is polar of P in relation to this conic,

$$RQ(x,y)\equiv (Ax+By+D)x_1+(Bx+Cy+E)y_1+(Dx+Ey+F)=0$$

On the other hand, the equation $\lambda(PR(x,y).PQ(x,y))+\mu(RQ(x,y))^2=0$ represents all conics which are touched by lines PR and PQ at points R and Q. Therefore, for especific values of $\lambda$ and $\mu $ (none of which can be equal to zero, because otherwise S would be a degenerate conic):

$$S(x,y)\equiv \lambda(PR(x,y).PQ(x,y))+\mu(RQ(x,y))^2=0$$

Then, $$S(x_1,y_1)=\lambda(PR(x_1,y_1).PQ(x_1,y_1))+\mu(RQ(x_1,y_1))^2,$$ $$S(x_1,y_1)=\mu(RQ(x_1,y_1))^2$$

Besides that,

$$RQ(x_1,y_1)=(Ax_1+By_1+D)x_1+(Bx_1+Cy_1+E)y_1+(Dx_1+Ey_1+F),$$ $$RQ(x_1,y_1)=S(x_1,y_1)$$


$$S(x_1,y_1)=\mu(S(x_1,y_1))^2,$$ $$\mu=\frac {1}{S(x_1,y_1)}$$


$$S(x_1,y_1).S(x,y)\equiv S(x_1,y_1)\lambda(PR(x,y).PQ(x,y))+(RQ(x,y))^2,$$ $$S(x_1,y_1)\lambda(PR(x,y).PQ(x,y))\equiv S(x_1,y_1).S(x,y)-(RQ(x,y))^2$$

Finally, equating left and right members of this identity to zero, we get that the equation of tangents PR and PQ to conic S can be represented by equation



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