# Can a redundant condition be removed from a solution?

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?

• Normal will remove the conditions, e.g., if solCond is the conditional solutions, then sol = solCond // Normal is the desired result. Nov 11, 2023 at 14:50

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)]

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}]

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