# The equation of a pair of tangents to a circle from a point.

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.

• 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. – coproc Oct 18 '15 at 19:35
• I am talking about the equation of pair of tangents to circle through a point P. See iit-jee-maths.blogspot.in/2008/12/… – 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.

• 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. – G-man Oct 20 '15 at 6:10
• I did not follow the derivative-tangent argument. Why should the equality of the curve equation's derivative imply tangency? – Isomorphism Oct 20 '15 at 9:17
• @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. – G-man Oct 20 '15 at 15:38
• @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? – MathGeek Aug 28 '16 at 10:46
• How did you conclude that the given second degree equation represents a pair of straight lines only? – 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,

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

Thus

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

Therefore

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

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

$$S(x_1,y_1).S(x,y)-(RQ(x,y))^2=0$$