First: This question is a follow-up question to my question here.
Consider the following code with output:
ClearAll[a1, b1, a2, b2, xC1, yC1, xC2, yC2, c, m]
a1 := 2
b1 := 1
a2 := 1
b2 := 1/2
xC1 := 0
yC1 := 1
xC2 := 0
yC2 := 4
eqn1 = a1 (a1^2 b2^2 m^2 + b1^2 (b2^2 - (c + m xC1 - yC1)^2)) ==
b2 (a1^2 (a2^2 m^2 + b2^2) - a2^2 (c + m xC2 - yC2)^2);
eqn2 = (a1^2 m (c - yC1) - b1^2 xC1)^2 ==
(a1^2 m^2 + b1^2) (b1^2 xC1^2 + a1^2 (c - yC1)^2 - a1^2 b2^2);
Solve[eqn1 && eqn2, {c, m}]
\begin{equation} \small \left\{ \left\{c \to -2, m \to -\frac{\sqrt{35}}{2}\right\}, \left\{c \to -2, m \to \frac{\sqrt{35}}{2}\right\}, \left\{c \to 2, m \to -\frac{\sqrt{3}}{2}\right\}, \left\{c \to 2, m \to \frac{\sqrt{3}}{2}\right\} \right\} \end{equation}
I need to calculate the values of the two expressions
(a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2)
(a2^2 m (c - yC2) - b2^2 xC2)/(a2^2 m^2 + b2^2)
for all four combinations of $c$ and $m$. (This gives eight output values.)
Furthermore, I would like the eight output values to be named x1
to x8
.
How do I do that?
Answer
ClearAll[a1, b1, a2, b2, xC1, yC1, xC2, yC2, c, m]
a1 = 2;
b1 = 1;
a2 = 1;
b2 = 1/2;
xC1 = 0;
yC1 = 1;
xC2 = 0;
yC2 = 4;
eqn1 = a1 (a1^2 b2^2 m^2 + b1^2 (b2^2 - (c + m xC1 - yC1)^2)) ==
b2 (a1^2 (a2^2 m^2 + b2^2) - a2^2 (c + m xC2 - yC2)^2);
eqn2 = (a1^2 m (c - yC1) - b1^2 xC1)^2 == (a1^2 m^2 + b1^2) (b1^2 xC1^2 +
a1^2 (c - yC1)^2 - a1^2 b2^2);
sol = Solve[eqn1 && eqn2, {c, m}];
Clear[x]
Using an indexed variable and depending on the order of the sequencing desired, use either
Evaluate[Array[x, 8]] = {(a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2),
(a2^2 m (c - yC2) - b2^2 xC2)/(a2^2 m^2 + b2^2)} /. sol // Flatten;
x /@ Range[8]
(* {Sqrt[35]/6, Sqrt[35]/3, -(Sqrt[35]/6), -(Sqrt[35]/3), -(Sqrt[3]/
2), Sqrt[3], Sqrt[3]/2, -Sqrt[3]} *)
or
Clear[x]
Evaluate[Array[x, 8]] = {(a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2),
(a2^2 m (c - yC2) - b2^2 xC2)/(a2^2 m^2 + b2^2)} /. sol // Transpose //
Flatten;
x /@ Range[8]
(* {Sqrt[35]/6, -(Sqrt[35]/6), -(Sqrt[3]/2), Sqrt[3]/2, Sqrt[35]/3, -(Sqrt[35]/
3), Sqrt[3], -Sqrt[3]} *)
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