Let the two circles C', C'' be given, and assume that a point O is the center of a circle C that is orthogonal to both C' and C'' (intersecting at points X and X' respectively). The tangents to C' and C'' at X and X' are normal to C, so they both pass through O. Since X and X' are both on C, OX and OX' are of the same length. On the other hand, if you had a point P on the plane outside C and C' such that if you drew tangents to C and C' and those distances PX and PX' were the same length, then it is obvious that the circle with center P and radius PX is orthogonal to both C and C'. Therefore, the loci we are looking for is all of the P described above. Since it wasn't immediately familiar to me, I derived a formula for the distance from a circle with center O = (x',y') and radius r to a point P = (x,y) outside the circle along a tangent that intersects the circle at T. OTP is a right angle, so a quick application of the Pythagorean Theorem gives us that PT = sqrt((x-x')^2 + (y-y')^2 - r^2). It goes without saying that the distance is independent of which of the two tangents is chosen. Let our two circles C, C' be given. Since we can choose our origin, axis, and scale without loss of generality, assume that C is centered on the origin (0,0) with radius 1, and C' is centered at (x',0) with radius r. Let a point P = (x,y) be an arbitrary point that is the center of a circle orthogonal to C and C'. The distance from P to C along the tangent line is sqrt(x^2 + y^2 - 1) and from P to C' is sqrt((x-x')^2 + y^2 - r^2). Since these two distances are to be equal, we square them both to get x^2 + y^2 - 1 = (x-x')^2 + y^2 - r^2 -2x'x = x'^2 - r^2 + 1 x = (x'^2 - r^2 + 1)/ -2x' So the curve is a straight line that is perpendicular to the line connecting the centers of the two circles, or rather that portion of the straight line that is not inside either of the circles.
Note that if the original circles intersect, the line in question passes
through the points of intersection of the two circles.