Question

Let a line 3x plus y equals to 1, and there are two circles made which touch the given line at points (-2, 7), and (2, -5). Given the locus of point of intersection of two circles drawn through the two points respectively.

Answer 1

$3x + y - 1 = 0$

Answer 2

Approach:

1. Equation of any circle touching the line
   A circle touching the line $3x + y - 1 = 0$ at a fixed point $(x_1,y_1)] has equation

   $$(x - x_1)^2 + (y - y_1)^2 = \lambda\,(3x + y - 1)^2,$$

   where (\lambda$ is a parameter.

2. Family through ((-2,7)] and another through ((2,-5)]
   Let the two circles be

   $$(x + 2)^2 + (y - 7)^2 = \lambda\,(3x + y - 1)^2, \quad (x - 2)^2 + (y + 5)^2 = \mu\,(3x + y - 1)^2.$$

3. Common points
   Any intersection ((x,y)] of these two circles must satisfy both equations. Subtracting one from the other eliminates the quadratic term in ((3x+y-1)^2] and leads to a linear condition in (x, y].

4. Derivation of locus
   Upon simplification, the locus collapses exactly to

   $$3x + y - 1 = 0.$$

Key Result:
The set of all common intersection points (other than the given touching points) lies on the given line itself.