Coordinates of the intersection of two tangents to a circle

Question

Let $A = (x_A, y_A)$ and $B = (x_B, y_B)$. Let $\gamma$ be a circumference of radius $r$, centered in $(0, 0)$; $A$ and $B$ lie outside of $\gamma$, and on the same side of some line $L$ through the center of $\gamma$.

Make tangents from $A$ and $B$ to $\gamma$. These tangents intersect in a point $C$, on the same side. What are the coordinates of $C$?

In green, you can see the tangents from $A$ and from $B$, which intersect in $C$.

I could find a solution by finding the equation of each tangent, and then finding their intersections. However, such a solution involves manipulating quite a few parametric polynomials, which takes a long time and is rather inelegant. Is there a simpler solution relying upon geometrical theorems?

Answer 1

To begin with we may assume the circle $C$, center $(0,0),$ has radius $1$ (divide all coordinates by the circle radius, then remultiply at the end). Rewrite each of $A,B$ in polar coordinates as $(r_a,\theta_a),\ (r_b,\theta_b).$ For each of $A,B$ there are then two points on $C$ where the tangents from $A$ or $B$ meet $C$, say they are at $t_a,t_b$ in the sense that their coordinates are e.g. $P(t_a)=(\cos(t_a),\sin(t_a))$ for $A$ and similarly for $B.$ The values for these $t$ are $$t_a=\theta_a \pm \cos^{-1}(1/r_a) \tag{1}$$ and similarly for $t_b$, as may be seen from a right triangle with one vertex at $A$, another at $P(t_a)$ and the third at the origin $O=(0,0).$ [right angle is at $P(t_a).$] The $\pm$ choice reflects which of the two tangents from $A$ one wants to use.

Once we have made choices for which tangent to use at each point, which amounts to choosing the sign in $(1),$ we have specific values for the angles $t_a$ and $t_b.$ Then the polar coordinates for the intersection point of the two tangents has its angle as $(t_a+t_b)/2$ and its radius as $\sec[(t_a-t_b)/2].$

Answer 2

Let $L$ be the tangent line from $A(s, t)$ to the circle (γ) touching it at $T$ (and also $T’$ too) with slope = $m$ and angle of inclination = $θ$.

Equation of $L$ is given by $y – t = m(x – s)$

Solving $L$ and (γ), with the added condition of equal roots for tangency, we found $m$ (after choosing the right one) and hence $θ$.

Let $L$ cuts the $x$-axis at $A’(a, 0)$, for some $a$ to be determined.

Then, $0 – t = m(a – s)$ giving $a = \frac {–t}{m} + s$.

In the figure, $\angle BCO = \angle A'CO = \angle COA’ = (0.5)θ$

Note also that $⊿A’OC$ is isosceles with $A’O = A’C = a$. Together with $\angle CA’O = π – θ$, $OC$ can be found by cosine law.

∴ $C$ is located at $(OC, (0.5)θ)$ in polar co-ordinates.

Note-1. According to the question, m will never be zero, otherwise there will be infinitely many or no C.

Note-2. If θ = 90 degrees, things should be treated separately but then (0.5)θ = 45 degrees.

Note-3. If A is on the x-axis, the proof can be further simplified. I think, through simple adjustments, such assumption is allowable.