0
$\begingroup$

A solution to this equation:

Solve[{(p-x)^2 + (q-y)^2==d, p/q==t}, {p,q}, Reals]

has this repetitive condition:

$\displaystyle\quad d > \frac{x^2 - 2txy+t^2y^2}{1+t^2}$

Here is the solution in question:

                                                               2    2              2  2         2              2  2
                                         t x + y        d + d t  - x  + 2 t x y - t  y         x  - 2 t x y + t  y
Out[45]= {{p -> ConditionalExpression[t (------- - Sqrt[-------------------------------]), d > --------------------], 
                                              2                          2 2                               2
                                         1 + t                     (1 + t )                           1 + t
 
                                                       2    2              2  2        2              2  2
                                 t x + y        d + d t  - x  + 2 t x y - t  y        x  - 2 t x y + t  y
>     q -> ConditionalExpression[------- - Sqrt[-------------------------------], d > --------------------]}, 
                                      2                          2 2                              2
                                 1 + t                     (1 + t )                          1 + t
 
                                                          2    2              2  2         2              2  2
                                    t x + y        d + d t  - x  + 2 t x y - t  y         x  - 2 t x y + t  y
>    {p -> ConditionalExpression[t (------- + Sqrt[-------------------------------]), d > --------------------], 
                                         2                          2 2                               2
                                    1 + t                     (1 + t )                           1 + t
 
                                                       2    2              2  2        2              2  2
                                 t x + y        d + d t  - x  + 2 t x y - t  y        x  - 2 t x y + t  y
>     q -> ConditionalExpression[------- + Sqrt[-------------------------------], d > --------------------]}}
                                      2                          2 2                              2
                                 1 + t                     (1 + t )                          1 + t

So I made that condition into an assumption:

Assuming[d > (x^2 - 2*t*x*y+t^2*y^2)/(1+t^2), {Solve[{(p-x)^2 + (q-y)^2==d, p/q==t}, {p,q}, Reals]}]

but it does not influence on the solution, i.e. it does not even change.

Question: is there a method of removing a condition made redundant by an assumption?

$\endgroup$
1
  • 1
    $\begingroup$ Normal will remove the conditions, e.g., if solCond is the conditional solutions, then sol = solCond // Normal is the desired result. $\endgroup$
    – Bob Hanlon
    Nov 11, 2023 at 14:50

2 Answers 2

1
$\begingroup$

The simplest way, no pun intended, to remove your condition, tell Simplify it is true.

Simplify[Solve[{(p-x)^2+(q-y)^2==d,p/q==t},{p,q},Reals],d>(x^2-2 t x y+t^2 y^2)/(1+t^2)]
$\endgroup$
0
$\begingroup$

Simply remove Reals from your code if you do not want the condition.

Solve[{(p-x)^2 + (q-y)^2==d, p/q==t}, {p,q}]

$$\left\{\left\{p\to -\frac{t \sqrt{d t^2+d+t^2 \left(-y^2\right)+2 t x y-x^2}}{t^2+1}+\frac{t^2 x}{t^2+1}+\frac{t y}{t^2+1},\\q\to \frac{-\sqrt{d t^2+d+t^2 \left(-y^2\right)+2 t x y-x^2}+t x+y}{t^2+1}\right\},\left\{p\to \frac{t \sqrt{d t^2+d+t^2 \left(-y^2\right)+2 t x y-x^2}}{t^2+1}+\frac{t^2 x}{t^2+1}+\frac{t y}{t^2+1},\\q\to \frac{\sqrt{d t^2+d+t^2 \left(-y^2\right)+2 t x y-x^2}+t x+y}{t^2+1}\right\}\right\}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.