2 equations and 2 unknown seemed easy to solve but not.

ucanbizon

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Hi,

I hope I am at the correct part of this forum to ask following question.

I have 2 equations with two unkown (x and y)

A=(b-x)^2 + (d-y)^2
F=(h-x)^2 + (k-y)^2

All variables except x,y is known and I want to calculate x and y values.

It seemed me easy at first because I had 2 equations and 2 unknown. But I saw that it is beyond my math level.

How can I solve it?

Thanks in adnavce
 
Hi,

I hope I am at the correct part of this forum to ask following question.

I have 2 equations with two unkown (x and y)

A=(b-x)^2 + (d-y)^2
F=(h-x)^2 + (k-y)^2

All variables except x,y is known and I want to calculate x and y values.

It seemed me easy at first because I had 2 equations and 2 unknown. But I saw that it is beyond my math level.

How can I solve it?

Thanks in advance
These are non-linear simultaneous equations.

These are two ellipses - which may intersect at 0, 1, 2, 3 or 4 points (solutions).

What methods have you been taught to solve such equations?
 
These are non-linear simultaneous equations.

These are two ellipses - which may intersect at 0, 1, 2, 3 or 4 points (solutions).

What methods have you been taught to solve such equations?

Thank you for your kind answer.I have converted the first equation to x=b-sqr(A-(d-y)^2)

and substitute the x in the second equation to convert second equation to something y= ....... with only known variables at the right side of y=....

But I saw that it is not so smple to get y=..... equation.
 
These are non-linear simultaneous equations.

These are two ellipses - which may intersect at 0, 1, 2, 3 or 4 points (solutions).

What methods have you been taught to solve such equations?

I have leaved the x alone at first equation ->x=b-sqr(A-(d-y))

And tried to substitute it at the second equation.

At first I have thought that I can handle it because there were 2 equation and 2 unknown. How can it be diffucult? But saw that I am in a deep problem. How can I slove it I have no idea.

By the way, this is my second anwser to you. I don't know where did the first submit gone.
 
Changing your equations to (same thing!):
(a-x)^2 + (b-y)^2 = u
(c-x)^2 + (d-y)^2 = v
Expand:
a^2 - 2ax + x^2 + b^2 - 2by + y^2 = u
c^2 - 2cx + x^2 + d^2 - 2dy + y^2 = v
Subtract:
a^2 - c^2 - 2ax + 2cx + b^2 - d^2 - 2by + 2dy = u - v
x and y together:
x(c - a) + y(d - b) = (u - v - a^2 + c^2 - b^2 + d^2) / 2

Since they're givens, let:
e = c - a
f = d - b
g = (u - v - a^2 + c^2 - b^2 + d^2) / 2
So we have:
ex +fy = g

Looks like all you can do is solve for x if y is given, or for y if x is given...
Denis, wait a minute. I am not exactly saying you're wrong (although I'd love to) but I do have a problem (actually many). You have the equation of two circles and your answer says that there are infinitely many points of intersections (all the points on the line ex+fy=g). What is going on here?!
 
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Denis, wait a minute. I am not exactly saying you're wrong (although I'd love to) but I do have a problem (actually many). You have the equation of two circles and your answer says that there are infinitely many points of intersections (all the points on the line ex+fy=g). What is going on here?!

I tried it out with actual numbers for circles that I know intersect at two points. I didn't want to work the algebra all the way through, but I think that if you take the equation of a line y(x) that you get, and plug it back into one of the quadratics, you get a solution for the two x-coordinates of the intersection points.
 
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Hi,

I hope I am at the correct part of this forum to ask following question.

I have 2 equations with two unkown (x and y)

A=(b-x)^2 + (d-y)^2
F=(h-x)^2 + (k-y)^2

All variables except x,y is known and I want to calculate x and y values.

It seemed me easy at first because I had 2 equations and 2 unknown. But I saw that it is beyond my math level.

How can I solve it?

Thanks in adnavce

what are you trying to find?
"I want to calculate x and y values"
do you mean you want to calculate common x and y values, i.e. the intersection of the two graphs?
 
WolframAlpha

An expression for x or y is pages long. However a specific numerical solution is easily obtained by graphing the two circles and reading the approximate intersections.

Another way to get a solution (exact and approximate) is WolframAlpha. Click below to see an example. You can use this WolframAlpha example by changing the constants' values in this example.

Click Here


(select "Approximate forms" to get decimal digits)
 
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Thank you very much indeed to all of you for your time to my qyestion.

Looks like all you can do is solve for x if y is given, or for y if x is given...

x and y are unknown and I need to calculate x or y first and substitute it in one of the equations to get other unknown (x or y).

what are you trying to find?
"I want to calculate x and y values"
do you mean you want to calculate common x and y values, i.e. the intersection of the two graphs?

Yes, I have two equations with unknown (x,y). All I need to find (x,y) solution set. I tried to convert equation 1 to something like x=........ (without y obviously at the right side of the equation)
And after calculating the x, subtitute it ate other equtaion to find y. But I am stuck. I have noticed that it is not as easy as it looks.

Another way to get a solution (exact and approximate) is WolframAlpha. Click below to see an example. You can use this WolframAlpha example by changing the constants' values in this example.

Thank you very much for the link. But it solves the equation (x,y) sets. But It doens't show how it is done. I am developing a software I I need to code it in my app. So I need to lear the way to handle such expressions.
 
I have 2 equations with two unkown (x and y)

A=(b-x)^2 + (d-y)^2
F=(h-x)^2 + (k-y)^2

All variables except x,y is known and I want to calculate x and y values.

Let's first rewrite the equations using a more typical form for circles; note that A and F are the squares of the radii of the circles, which I'll call r^2 and s^2, and the centers are (b,d) and (h,k). The equations are then

(x-b)^2 + (y-d)^2 = r^2
(x-h)^2 + (y-k)^2 = s^2

Here is a place where I solved essentially the same problem 16 years ago, though I didn't show all the details of my work; the key to keeping it simple was to use intermediate variables to control the complexity of the equations. There is also a link to a page that gives a computer-generated solution in all its ugliness. I imagine the same solution as mine could also be obtained by geometric rather than algebraic means.
 
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