I'm attempting to follow the accepted solution here:
However, I cannot seem to get $k$ to equal out to the correct scalar value.
In my use case I made up a triangle of the following 3 points ${(0,0),(921,0), (460.5,265)}$ and I want to reduce the radius by 25.
Here is what I have so far:
$a=\sqrt((921-460.5)^2+(0-265)^2)=531.305$
$b=\sqrt((0-460.5)^2+(0-265)^2)=531.305$
$c=\sqrt((0-921)^2+(0)^2)=921$
$P = a+b+c = 1983.61$
$A = (.5)(921)(265)=122032.5$
$r = 2*A/P=123.041$
$Q = c/2,r = (460.5,123.041)$
$k=r/(r-x) = 123.041/(123.041-(-25))= 0.831$
Solving for points A and C
$A’ = A+(Q-A) * k$
$Xa=0+(460.5-0)*.831 = 382.676$ Is this a vector from the inscribed center back towards 0? $Ya=0+(123.041-0)*.831 = 102.247$
$Xc=460.5+(460.5-460.5)x.831 = 460.5$
$Yc=265+(123.041-265)x.831= 147.197$
Would the inverse of k be $k'= (r+x)/r$ so if I wanted r to grow by 200 k would be
$k = (123.041+200)/123.041 = 2.621$?
I would comment in that question, but I do not have the reputation to do so, sorry for asking a question related to a solved problem but I'm really just not understanding the solution.
Could someone please show me how to arrive at the correct solution.
I think I found the answer and the problem exists in the original question.
$k = R/(R-X)$
This is not accurate. Using the numbers above for points on the x-axis where y is 0 and solving for k using $A'=A+(Q-A)*k$ would give the following:
$25=0+(123.041-0)*k$
$k = 25/123.041 = .203$ which is just $X/r$