The locus is the circle with diameter AB, except for points A and B.
This is a theorem that one set equals another, so that means that it is an if and only if theorem that requires the proof one a statement and also the converse statement.
In this case, triangle ACB is similar to triangle to B'CA' but NOT similar to triangle A'CB'. The proof is by SAS by checking the equal ratios: CB'/CA = CA'/CB. This ratio = 3/ab = k the scaling factor, so also k = A'B'/7. This shows A'B' = 21/ab.
Note: The letters a and b are numbers, so they are given as any possible numbers and stay the same throughout the problem. You cannot change them or solve for them.
There are two circles to be constructed. The centers of the circles are the intersections of the line through B perpendicular to line m with the two lines that bisect the angles formed by m and n (i.e., the locus of points equidistant from m and n). One the centers are constructed, since the circles are drawn to pass through B.
This problem is related to
This problem is a combination of thales figures (dilations) with transversals. The answers are (a) 4/13 (dilation, then transversal), (b) 4/13 (dilation) (c) 4/17 (transversals and addition of segments) (d) 1 (problem 3.3 of Assignment 3A).
There are many examples of such ratio arguments in B&B, pp. 64-65 and the problems about constructing ratios such as problem 3.7 (Angle Bisector Ratio Theorem) of Assignment 3B.
Also see Lab 5 (Dilations) and Trapezoid ratio interactive page linke to Week 4.