Consider a circle with center at origin and a fixed point $A=(a_1,a_2)$ lying on it (I think these two conditions describe the circle completely and uniquely because center is known and radius is known too as $\sqrt{a_1^2+a_2^2}$ ). Now, at $A$ a tangent is drawn to the circle. Picture up to now gonna look like this,
Now, consider an arbitrary point $P=(x_1,x_2)$ on the circle, therefore, $y_1=\sqrt{r^2 - x_1^2}$ where $r^2 = a_1^2 +a_2^2 $. Draw a tangent at $P$ and let this this tangent meet the tangent at $A$ in $T$. Join $A$ to $P$. Picture up to now gonna look like this,
We can see that $AP$ is the chord of contact, and theorem
If $T$ is a point with coordinates $(x',y')$ and from it two tangents are drawn to a circle $x^2 + y^2 = r^2$ then the equation to the chord of contact is $$x'x + y'y = r^2~~~~~~~~~~~~~~~~(1)$$
We can write the equation to $AP$ as $$ \frac{y-a_2}{x-a_1} = \frac{y_1-a_2}{x_1-a_1}$$ and doing some little rearrangements we have $$(a_2 - y_1)x + (x_1 - a_2)y = a_2x_1 - a_1y_1$$ If we compare it with equation (1) then we can say (as far as I think but according to me this is the real unsure thing) $$ a_2x_1 - a_1y_1 = r^2$$ $$ a_2x_1 -a_1\sqrt{r^2 - x_1^2} = r^2$$ $$ a_1^2(r^2 - x_1^2) = (r^2 - a_2x_1)^2$$ $$ a_1^2r^2 - a_1^2x_1^2 = r^4 + a_2^2x_1^2 - 2r^2a_2x_1$$ $$ 0 = x_1^2(a_1^2 +a_2^2) - 2r^2a_2x_1 + r^4 - a_1^2r^2 ~~~~\rightarrow 0= x_1^2r^2 - 2r^2a_2x_1 +r^4 - a_1^2r^2$$ (since $a_1^2 +a_2^2 = r^2$) $$ x_1^2 -2a_2x_1 +r^2 - a_1^2 =0$$ Solving this and using the fact that $a_1^2 + a_2^2 = r^2$ we would get $x_1 = a_2$ and consequently $y_1 = a_1$. My problem is that I considered $(x_1,y_1)$ to be completely arbitrary but they came out to be constants. What does this tell us? Doesn't it say that tangent at $(a_1,a_2)$ can be intersected only by a tangent at $(a_2,a_1)$ ?
P.S. :- This situation has arrived during the solution of the this problem:
The tangent at any point P of a circle meets the tangent at a fixed point A in T, and T is joined to B, the other end of the diameter through A; prove that the locus of the intersection of AP and BT is an ellipse whose eccentricity is $\frac{1}{\sqrt2}$.
I couldn't solve this question even now (although mentioning this is completely immaterial here).
I’m going to use $(x_1,y_1)$ and $(x_2,y_2)$ for the two points on the circle for consistency of notation. Your equation of the line through these two points then becomes $$(y_2-y_1)x+(x_1-x_2)y = x_1y_2-x_2y_1.\tag2$$ So far, so good. However, you can’t assume that this equation is identical to equation (1): they might be multiples of each other. This means in particular that you can’t conclude that the right-hand side of the above equation is equal to $r^2$.
At this point, if you’re trying to verify equation (1), you’ll need to compute the coordinates of $T$, plug them into (1) and show that the two equations define the same line. Omitting the details of the calculations, $$T={r^2\over x_1y_2-x_2y_1}\left(y_2-y_1,x_1-x_2\right).$$ This seems plausible. If the two points are the ends of a diameter, the above denominator vanishes, which corresponds to the two parallel tangents having no (finite) intersection. If you plug this into equation (1), rearrange and remove the common factor of $r^2$, you should eventually end up with equation (2).
As far as the original problem that you were trying to solve goes, I suggest working with the unit circle with $A=(1,0)$. Instead of using a point $P$ on the circle, you can then parameterize $T$ as $(1,t)$ and use formula (1) for line $AP$. An equation for $BT$ is easy to find, as is the intersection for the two lines. Eliminate $t$ from the resulting expression. Any other choice of circle and point $A$ is obtained from this via a similarity, which leaves the eccentricity of a conic invariant.