One equation equivalent to two equations

Question

Two circles can intersect. I want to find one equation with a set of solutions equivalent to two equations. $$o_1:x^2-2a_{1}x+{a_1}^2+y^2-2b_{1}y+{b_1}^2-{r_1}^2=0\\ o_2:x^2-2a_{2}x+{a_2}^2+y^2-2b_{2}y+{b_2}^2-{r_2}^2=0$$ Now addition gives an another circle. Subtraction gives a line – if I subtract zero from two sides of $o_1$ equation and do the same on $o_2$ equation, I have got : $$o_1:(-2a_{1}+2a_{2})x+(-2b_{1}+2b_{2})y+{a_1}^2-{a_2}^2+{b_1}^2-{b_2}^2-{r_1}^2+{r_2}^2=0\\ o_2:(-2a_{2}+2a_{1})x+(-2b_{2}+2b_{1})y+{a_2}^2-{a_1}^2+{b_2}^2-{b_1}^2-{r_2}^2+{r_1}^2=0$$ I thinked I can add to both side anything. Result is one equation that is by no means equivalent to the system of equation I staryed with. Can you point my fault ?

Answer 1

Whatever the transformation you apply, the original equations remain. You thought from

$$\begin{cases}o_1\\o_2\end{cases}$$ (two circles) to $$\begin{cases}o_1-o_2\\o_2-o_1\end{cases}$$ (twice the same straight line),

but in fact your system is now

$$\begin{cases}o_1\\o_2\\o_1-o_2\\o_2-o_1\end{cases}$$ (the two circles and twice the line).

The solutions are still two, one or no intersection points.

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Note that any linear combination of the initial equations gives you the equation of a circle (or exceptionally a straight line), that goes through the same common intersection.