Solving a system of nonlinear (quadratic) equations

Question

Consider the following system of equations: $$\begin{align} (x + 1)^2 [(p - l)^2 + (q - m)^2] &= (a - l)^2 + (b - m)^2 \\ (x + 1)^2 [(p - a)^2 + (q - b)^2] &= x^2[(a - l)^2 + (b - m)^2]\end{align}$$ Find the value of $p$ and $q$ in terms of $a, b, x, l$ and $m$, provided that $x > 0$.

Answer 1

Hint: Divide both equations by $(x+1)^2$ and set $$\alpha^2 = \frac{(a - l)^2 + (b - m)^2}{(x+1)^2} \quad \beta^2 = \frac{x^2[(a - l)^2 + (b - m)^2]}{(x+1)^2},$$ then the system of equations to solve becomes $$\left\{\begin{array}{rcl} (p - l)^2 + (q - m)^2 &=& \alpha^2 \\ (p - a)^2 + (q - b)^2 &=& \beta^2 \end{array}\right.$$ I.e. you have to find the intersection of two circles in the plane. One of these circle has center $c_1 = (l,m)$ and radius $r_1 = \alpha$ while the other has center $c_2 = (a,b)$ and radius $r_2=\beta$.

Answer 2

The right hand sides just differ by a factor $x^2 > 0$: $$ \begin{align} (x + 1)^2 [(p - l)^2 + (q - m)^2] &= (a - l)^2 + (b - m)^2 \\ (x + 1)^2 [(p - a)^2 + (q - b)^2] &= x^2[(a - l)^2 + (b - m)^2] \end{align} $$

so we can get an equation for the left hand sides: $$ \begin{align} (x + 1)^2 [(p - l)^2 + (q - m)^2] &= \frac{(x + 1)^2}{x^2} [(p - a)^2 + (q - b)^2] \iff \\ [(p - l)^2 + (q - m)^2] &= \frac{1}{x^2} [(p - a)^2 + (q - b)^2] \quad (*) \end{align} $$

and indeed if we call the left side of $(*)$ $r_1^2$ and $x^2$ times the right side $r_2^2$ we got two circle equations in coordinates $p$ and $q$ $$ \begin{align} (p - l)^2 + (q - m)^2 &= r_1^2 \\ (p - a)^2 + (q - b)^2 & = r_2^2 \end{align} $$

where $r_1 = \sqrt{(a-l)^2+(b-m)^2}\, / \, (1 + x)$, $r_2 = x \, r_1$ and the first circle has radius radius $r_1$, origin $(l, m)$ and the second circle has radius $r_2$, origin $(a, b)$.

So the system can have the following solutions:

• no solution: the two circles do not intersect, e.g. both origin $(0,0)$ and $r_1 \ne r_2$
• one solution: the two circles intersect at just one point
• two solutions: the two circles intersect in two points
• infinite solutions: the two circles are congruent (same origin, same radius)

This gives $$ p = l \pm \sqrt{r_1^2 - (q -m)^2} = a \pm \sqrt{r_2^2 - (q - b)^2} \quad (**) $$ For $l = a$ this reduces to $$ r_1^2 - (q -m)^2 = r_2^2 - (q - b)^2 $$ which gives $$ q = \frac{b^2-m^2+r_1^2-r_2^2}{2(b-m)} $$ for $b \ne m$ otherwise this reduces to $$ r_1 = r_2 \iff x = 1 $$ for $b = m$.

For $l \ne a$ the calculation gets lengthy, throwing it into a computer algebra system (here WolframAlpha) gives this result.

Once one knows $q$ one can use one of the equations $(**)$ to calculate $p$.

Answer 3

Here's my semi-geometrical solution. Let's scale $p$ and $q$ to $1+x$. So $$ P = p(1+x) \\ Q = q(1+x) $$ so equation become \begin{align} \left [P-(x+1)l \right ]^2 + \left [Q-(x+1)m \right ]^2 &= (a-l)^2 + (b-m)^2 \\ \left [P-(x+1)a \right ]^2 + \left [Q-(x+1)b \right ]^2 &= x^2 \left [ (a-l)^2 + (b-m)^2 \right ] \end{align} So, geometrical part now. The problem above is to find intersection points of two circles. Obviously circles might intersect at zero points (don't intersect), one point (tangent to each other), two points (real intersection) and infinite number of points (coinciding circles). Circle number one has a center at $[(x+1)l, (x+1)m]$ and a radius of $r = \sqrt{(a-l)^2 + (b-m)^2}$. Circle number two has a center of $[(x+1)a, (x+1)b]$ and a radius of $R = \sqrt{x^2 \left [ (a-l)^2 + (b-m)^2 \right ]} = rx$. Now, observe that distance between centers is $d = \sqrt{(x+1)^2 \left [ (a-l)^2 + (b-m)^2 \right ]} = (x+1)r = r + R$. So, the only possibility to those two circles is to be tangent to each other. Diagram is below

It is obvious now, that tangent point divides the line that connects the centers in $1:x$ proportion. So $$ P = (x+1)l + 1 \cdot (a-l) = a + xl \\ Q = (x+1)m + 1 \cdot (b -m) = b + xm $$ Going back to $p$ and $q$, you can easily find that $$ \left \{ \begin{array}{ccc} p & = & \frac {a + xl}{1 + x} \\ q & = & \frac {b + xm}{1 + x} \end{array} \right . $$