Length of tangent circles - locus I have two circles that their equations are
$$x^2+y^2-6x-27=0$$
$$x^2+y^2+2x-8 = 0$$
I need to find the line-locus of all the points that the length of the tangent from the first circle and the second are equal.
I would want actually a hint because I don't know how to start! The slope of the tangents and radiuses to the tangent always change.
 A: @Issac Browne It is a classical locus : it is a line called the radical axis of the 2 circles (https://en.wikipedia.org/wiki/Radical_axis). Using the concept of power of a point with respect to a circle, its equation is found by equalizing the two equations :
$$x^2+y^2-6x-27=x^2+y^2+2x-8,$$
giving a straight line orthogonal to the axis defined by the centers of the circles, with equation :
$$8x=-19 \ \ \iff \ \ x=\tfrac{-19}{8}.$$
A: The angle between the radius and the tangent line is a right angle, and we can calculate the radius as well as the distance from a point $(a,b)$ to the center of the circle. Thus we can also calculate the distance to the tangent point. Set the distance of the first one equal to the distance of the second one and there you have your equation! Note that it is a straight line. This locus which you have found is called the radical axis of a circle.
A: A circle’s center $C$, point external to the circle $P$ and point $T$ at which a line through $P$ is tangent to the circle form a right triangle with the right angle at $T$. So, for a given distance $d=PT$, you can use the Pythogorean theorem to find $PC$. Thus, for any given distance $d$, finding the points that have equal tangent lengths to your two circles amounts to finding the intersections of a pair of circles. Use this unknown distance $d$ as a parameter of the locus.
