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How to calculate the equation with letter and variable

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Dear everyone,
I want to calculate an equation with letter and variable.
Now I can get the variable value presented with the letters I used
but I don't know how to change the letters as real numbers.
this is the example
%declear syms
syms x0 y0 x1 y1 x2 y2 a b p q d positive;
[x0 y0] = solve('(x-p)^2+(y-q)^2=d^2','(p-a)*(y-b)=(x-a)*(q-b)')
then I can get the result
but the problem is how to enter the real number value of letters like
a = 1;
b = 2;
p = 3;
q = 4;
d = 5;
then I can get the numerical value of x, y ???

  1 Comment

HONG CHENG
HONG CHENG on 4 May 2017
Maybe we can use the subs() function to do this
f=subs(solve('(x-p)^2+(2-q)^2=d^2') ,{a,b,p,q,d},{1,2,3,4,5})
but here we just can use one equation.
if anyone have other or better method, please show your idea

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Accepted Answer

Star Strider
Star Strider on 4 May 2017
The substitution will occur automatically. The problem is that you must put ‘x’, ‘x0’ and ‘y0’ in the equations you want to solve for them.
This will do the substitutions:
syms x0 y0 x1 y1 x2 y2 a b p q d positive
a = 1;
b = 2;
p = 3;
q = 4;
d = 5;
Eq1 = (x-p)^2+(y-q)^2==d^2;
Eq2 = (p-a)*(y-b)==(x-a)*(q-b);
[x0 y0] = solve(Eq1, Eq2);

  9 Comments

Show 6 older comments
Star Strider
Star Strider on 6 May 2017
As always, my pleasure!
It appears that you have resolved this, and you have a method to choose the ‘correct’ ‘(x0,y0)’.
Walter Roberson
Walter Roberson on 7 May 2017
HONG CHENG comments on Star Strider's Answer:
kind and big god in MATLAB
Karan Gill
Karan Gill on 9 May 2017
See my answer below for why you only get one solution.

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More Answers (1)

Karan Gill
Karan Gill on 9 May 2017
Edited: Stephen Cobeldick on 17 Oct 2017
Use subs to substitute values, as shown below. You only get one solution because in the other solution, "x" is negative, which is not allowed due to the assumption that it is positive.
BUT if don't substitute values before solving, then you get two solutions because the second solution can be positive under certain conditions. "solve" also issues a warning stating that conditions apply to the solutions. If you use the "ReturnConditions" option, then you get these conditions. Applying these conditions will let you find correct values. See the doc: https://www.mathworks.com/help/symbolic/solve-an-algebraic-equation.html.
syms x y a b p q d positive
eqn1 = (x-p)^2+(y-q)^2 == d^2;
eqn2 = (p-a)*(y-b) == (x-a)*(q-b);
vars = [a b p q d];
vals = sym([1 2 3 4 5]);
eqn1 = subs(eqn1,vars,vals);
eqn2 = subs(eqn2,vars,vals);
[xSol ySol] = solve(eqn1, eqn2)
xSol =
(5*2^(1/2))/2 + 3
ySol =
(5*2^(1/2))/2 + 4
Lastly, do not redeclare symbolic variables as doubles because you are overwriting them. So don't do this.
syms a
a = 1
Just do
a = 1
Or use "subs" to substitute for "a" in an expression
f = a^2
subs(f,a,2)
Karan (Symbolic doc)

  3 Comments

HONG CHENG
HONG CHENG on 9 May 2017
dear Karan, thanks for your kind answer.
With your explanation,I know the real reason I get two values. Thanks a lot!
Karan Gill
Karan Gill on 9 May 2017
Glad to hear that! If my answer was helpful, could you accept it so that others can find the reason.
HONG CHENG
HONG CHENG on 10 May 2017
Sorry, I will reopen another question. Can you answer in this link - a new question ?
Because I think Another Answer does a lot help for me too.
I will accept the answer in this new link. Thanks a lot !!!1

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