# Is it possible to get "*" operators in the results of W|A?

Posted 15 days ago
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 Hello, I solve implicit equations and sometimes the results are somewhat lengthy. (The question is about Wolfram Alpha)Example: solve ((x2-x0)^2+(y2-y0)^2) = ((x1-x0)^2+(y1-y0)^2)=y0^2 for x0,y0 (Calculating the center (x0,y0) of a circle through 2 points (x1,y1),(x2,y2) tangentially touching the x-axis)I use the resulting formulas in C++ and Pascal.The "*" multiplication operators are missing and I must add them manually or write a utility to do this. Is there a ready made solution for this?
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Posted 15 days ago
 Try InputForm[Solve[((x2-x0)^2+(y2-y0)^2)==((x1-x0)^2+(y1-y0)^2)==y0^2,{x0,y0}]] or InputForm[Simplify[Solve[((x2-x0)^2+(y2-y0)^2)==((x1-x0)^2+(y1-y0)^2)==y0^2,{x0,y0}]]] 
Posted 15 days ago
 Thank you, I am not sure this is the answer, because what I want are multiplication operators in the plain text result. In Wolfram Alpha I get for example as solution no. 4: x0 = (-sqrt(y1 y2 (x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2)) - x1 y2 + x2 y1)/(y1 - y2) and y0 = (2 x1 sqrt(y1 y2 (x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2)) - 2 x2 sqrt(y1 y2 (x1^2 - 2 x1 x2 + x2^2 + y1^2 - 2 y1 y2 + y2^2)) + x1^2 y1 + x1^2 y2 - 2 x1 x2 y1 - 2 x1 x2 y2 + x2^2 y1 + x2^2 y2 + y1^3 - y1^2 y2 - y1 y2^2 + y2^3)/(2 (y1 - y2)^2) and y1 - y2!=0 and y1!=0 If there where "*" operators, I could paste this code into my program, refactor it (if necessary) and use it directly. Not the length is my problem, this is unavoidable; The problem is I have to add all "*" operators.