Find the combined equation of two tangents drawn from $P(x_1,y_1)$ to the circle $x^2+y^2 = a^2$. Point $P$ lies outside the circle.
[Math] To find tangents to given circle from a point outside it
circles
Related Solutions
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
$$S(x_1,y_1).S(x,y)-(RQ(x,y))^2=0$$
Without the loss of generality say the circle is $x^2+y^2=r^2$. Let $P(a,b)$ be the external point. A line passing through $P$ with slope $m$ is of the form $$y-b=m(x-a).$$
To get the intersection points of this line and the circle, we substitute $y$ in the equation of the circle and obtain $$x^2(m^2+1)+2mx(b-am)+((am-b)^2-r^2)=0.$$
For tangency, we want equal roots. Thus we want the discriminant to be $0$. This gives us $$4m^2(b-am)^2-4(m^2+1)((am-b)^2-r^2)=0.$$ Upon simplifying we get, $$m^2r^2-(am-b)^2-r^2=0.$$ This is a quadratic equation in $m$ so at the most two real values. This means there can be only two possible tangent lines emanating from $P$.
In fact, it can be easily shown that if this equation has a real root then it will have two distinct real roots, thus exactly two tangents.
Best Answer
Sorry for the lack of a picture. You can show that the angle between the line from the origin to $(x_1,y_1)$ and a radial line at a point of tangency is
$$\tan{\theta} = \frac{\sqrt{x_1^2+y_1^2-a^2}}{a}$$
where $a$ is the radius of the circle centered at the origin. Let $\theta_1$ be the angle between the line from the origin to $(x_1,y_1)$ and the positive $x$ axis. Then the tangent points on the circle are given by
$$(a \cos{(\theta_1 \pm \theta)},a \sin{(\theta_1 \pm \theta)})$$
The combined equation of the tangent lines is then
$$y-y_1 = m_{\pm} (x-x_1)$$
where
$$m_{\pm} = \frac{y1-a \sin{(\theta_1 \pm \theta)}}{x_1-a \cos{(\theta_1 \pm \theta)}}$$