Point of Intersection of tangents Tangent are drawn to the circle $$x^2+y^2=1$$ at the point where it is met by the circles $$x^2+y^2-(\omega+6)x+(8-2\omega)y-3=0$$ where $$\omega$$  being the variable. Find the locus of point of intersection of these tangents.
 A: The circles meet on the line $$-(\omega+6)x+(8-2\omega)y-2=0$$
which is obtained subtracting the equations of the two circles.
The normal vector to this line is $(-\omega-6,8-2\omega)$. Therefore the point on that line closer to the origin is $$\frac{2}{(\omega+6)^2+(8-2\omega)^2}(-\omega-6,8-2\omega)$$
By Pythagoras or by taking into account that this point and the intersection of the tangents are inverses of each other with respect to the unit circle, the point of intersection of the tangents is this point divided by the square of its norm. 
$$\frac{1}{2}(-\omega-6,8-2\omega)=\frac{1}{2}(-1,-2)\omega+(-3,4)$$
As $\omega$ moves this travels a line pointing in the direction of $(-1,-2)$ and passing through $(-3,4)$.
A: This is a supplement to Marja’s answer, following the same derivation but using projective geometry methods for the calculations.
Subtracting one circle equation from the other, we get the line $$(\omega+6)x+(2\omega-8)y+2=0,$$ which we represent by the homogeneous vector $\mathbf l=[\omega+6:2\omega-8:2]$. (We could also have found this by subtracting the matrices of the two circles and extracting the line from the resulting singular matrix, but this is quicker.) The pole of the line through a pair of points on a conic is the intersection of the tangents through those points. The unit circle’s matrix is $C=\operatorname{diag}(1,1,-1)$, i.e., the equation of this circle can be written as $(x,y,1)C(x,y,1)^T=0$. This matrix is its own adjugate, so the intersection of the tangents at the points where $\mathbf l$ intersects $C$ is given by $$C^*\mathbf l=C\,\mathbf l=[\omega+6:2\omega-8:-2].$$ This line is the join of $[6:-8:-2]$ and $[1:2:0]$, which we can convert to non-parametric Cartesian form by taking their cross product, giving $[-4:2:-20]$, i.e., the line $$2x-y+10=0.$$
