You're given $\triangle ABC$, with known sides, find the location of point $P$ such that
$ PB = k_1 PA $ and $ PC = k_2 PA $ where $k_1 \gt 0, k_2 \gt 0$ are given constants.
So for this, I translated the given two conditions into
$ (P - B)^T (P - B) = k_1^2 (P - A )^T (P - A) $
$ (P - C)^T (P - C) = k_2^2 (P - A )^T (P - A ) $
Take the first of these two,
$ (1 - k_1^2) P^T P + 2 P^T ( k_1^2 A - B ) + B^T B - k_1^2 A^T A = 0 $
There are two cases to consider. If $k_1 = 1$ then we get the equation of a line perpendicular to $(A - B)$, and passing through $ \dfrac{1}{2} (A + B) $, i.e. the perpendicular bisector of $AB$.
If $ k_1 \ne 1 $ then we get a circle, with center $ \dfrac{ k_1^2 A - B }{k_1^2 - 1} $ which lies on the line $AB$.
The same can be said about the second equation. Hence the location of $P$ is the intersection of these two curves (which are either a circle or a line).
Example: Suppose $A = (0,0), B = (14, 0), C = (5, 12) $, and let $k_1 = 2 , k_2 = 3 $, then we have to find the intersection of the following two circles:
$ (-3) (x^2 + y^2) + 2 (x, y) \cdot ( 4 (0, 0) - (14, 0) ) + (196 - 4 (0) ) = 0 $
$ (-8) ( x^2 + y^2) + 2 (x, y) \cdot ( 9(0,0) - (5, 12) ) + ( 169 - 9(0) ) = 0 $
After re-arranging, these become
$3 x^2 + 3 y^2 + 28 x - 196 = 0$
$8 x^2 + 8 y^2 + 10 x + 24 y - 169 = 0 $
But these two circles do not intersect, so for these values of $k_1, k_2$, there is no such point $P$.
So I modified the values of $k_1, k_2$ to $ k_1 = 1.25 , k_2 = 1.5$
The equation of the circles become
$ - \dfrac{ 9}{16} (x^2 + y^2) + 2 (x,y) \cdot ( - (14, 0) ) + 196 = 0 $
$ - \dfrac{5}{4} (x^2 + y^2) + 2 (x, y) \cdot ( - (5, 12) ) + 169 = 0$
And these two equations produce two solutions
$ (x_1, y_1) = ( 6.140419 , 2.254615 ) $
$ (x_2, y_2) = (-6.39222, -25.0155) $
These are the intersections of the two circles, and only $(x_1, y_1)$ is within the triangle.
My question is: Can someone verify my findings, or suggest an alternative way to solve the problem ?