3
$\begingroup$

Problem:

Solve for the values of a, b

Equation 1:

$$(x_1-a)^2+(y_1-b)=r^2$$

Equation 2:

$$(x_2-a)^2+(y_2-b)^2=r^2$$

Where, $x_1, x_2, y_1, y_2$ and $r$ are all constant values

For the purposes of this problem $x_2 = 0$ and $y_1 = 0$

My Solution:

Eq. 1 and 2 become

**Equation 1a:**$$(x_1-a)^2+b^2=r^2$$ **Equation 2a:**$$a^2+(y_2-b)^2=r^2$$

Expanding Eq. 1a

**Equation 1b:**$$b^2+a^2-2ax_1-r^2+x_1^2=0$$ **Equation 2b:**$$a^2+b^2-2by_2-r^2+y_2^2=0$$

Solving for b $$b=\pm\sqrt{r^2-a^2+2ax_1-x^2}$$ Substituting the value of b into Eq. 2b and solving for a using the quadratic equation I end up with the coefficients to the quadratic equation as such.

$$A=(4x_1^2+4y_2^2)$$ $$B=(4x_1y_2^2-8y_2x_1-4x_1^3)$$ $$C=(2y_2^2x_1^2-4y_2^2r^2+x_1^4+y_2^4)$$

plugging in the coefficients into $$f(x)=\frac{-b^2\pm\sqrt{b^2-4ac}}{2a}$$

I then have the two values of a which I subsequently plug back into Eq. 1a and solve for b.

note: To test my solution I have used the equation of a circle with non-zero (a, b) values to determine what the value of a, b should be at various x and y values.

My Question(s):

The value I calculate for a is approximately half of the true value of a yet despite extensive review I cannot find my error.

1: Is there anything unique about solving a system of quadratic equations that would create error?

2: Is it likely that the error is a rounding error introduced by the program used to calculate a?

3: Is there something wrong with my approach?

Exact Solution:

Subtracting Eq. 2b from Eq. 1b

$$-2ax_1+2by_2+x_1^2-y_2^2=0$$ $$b=\frac{ax_1}{y_2}-\frac{x_1^2}{2y_2}+\frac{y_2}{2}$$

When b is substituted into Eq. 1b we get the following coefficients which can be plugged into the quadratic equation to get a.

$$A=(1+\frac{x_1^2}{y_2^2})$$ $$B=(-x_1-\frac{x_1^3}{y_2^2})$$ $$C=(\frac{x_1^2}{2}+\frac{x_1^4}{4y_2^2}+\frac{y_2^2}{4}-{r^2})$$

$\endgroup$
  • $\begingroup$ When you substitute the value of $b$ in equation $2b$, the term $a^2$ is canceled out. You can then obtain $a$ by a simple linear equation. $\endgroup$ – Anatoly Jul 14 '14 at 21:01
  • 2
    $\begingroup$ A better approach than yours is to "subtract" 2b from 1b. We end up with a linear equation in $a$ and $b$. Using that equation, solve for $b$ in terms of $a$ (can't in a few cases) and substitute in 1b. Now we have a quadratic, solve. $\endgroup$ – André Nicolas Jul 14 '14 at 21:02
  • 1
    $\begingroup$ The two distinct circles will intersect in either zero, one (if they're tangent), or two points. The linear equation André Nicolas describes is the line on which the two intersection points lie (for that third case). Inserting that relation into either circle equation will give you the coordinates for the intersections without ambiguity. $\endgroup$ – colormegone Jul 14 '14 at 21:51
  • $\begingroup$ It is necessary to solve something like this. Rewriting formula bit different. math.stackexchange.com/questions/757664/… $\endgroup$ – individ Jul 15 '14 at 13:16
  • $\begingroup$ @AndréNicolas your method simplified the calculation significantly. I will post the solution I got using this method above. Thanks. $\endgroup$ – MathUsiast Jul 15 '14 at 18:19
0
$\begingroup$

The system of equations:

$$\left\{\begin{aligned}&x^2+y^2=z^2\\&q^2+t^2=z^2\end{aligned}\right.$$

the solutions have the form:

$$x=4p^4-s^4$$

$$y=4p^2s^2$$

$$q=4ps(2p^2-s^2)$$

$$t=4p^4-8p^2s^2+s^4$$

$$z=4p^4+s^4$$

$p,s$ - integers.

Although it is written decision system Diofantovy equations. I think the difficulty is not to find solutions such what .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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