We have two non intersecting in $\mathbb R^2$ circles: $ C_1(x,y)\equiv x^2+y^2-2mx-2ny+p=0, $ $ C_2\equiv x^2+y^2-2m'x-2n'y+p'=0, $ where circles are with real coefficients and positive radiuses. Then they intersects in some $(a,b), (a',b')\in \mathbb C^2$. Let $C$ be an arbitrary circle, with or without real points, with real coefficients, passing throw $(a,b), (a',b')$. I wish to know, is $C$ is the form $\alpha C_1(x,y)+\beta C_2(x,y)$, for some real $\alpha, \beta$ ?
Thanks
On
The equation of a circle is
quadratic,
with equal coefficients of the squares, $x^2$ and $y^2$,
without mixed term $xy$.
Now the combination $\alpha C_1+\beta C_2$ is
quadratic,
with equal coefficients of the squares, $x^2$ and $y^2$,
without mixed term $xy$
hence it is the equation of a circle.
Plus, any point that belongs to both circles satisfies $\alpha C_1+\beta C_2=0$ and also belongs to that circle. So if the two given circles make two intersections, the whole family shares these intersections.
Notice that the coefficients of the squares can vanish, leaving the equation of a straight line as a degenerate case. For the same reason as below, this is the line through the two intersection points.
The answer is YES
Given $$C_1:x^2+y^2-2mx-2ny+p=0;\;C_2:x^2+y^2-2m'x-2n'y+p'=0$$ The linear combination
$$\mathcal{C}:\lambda C_1+\mu C_2=0$$ $\mathcal{C}$ gives the family of all circles, real or complex, passing through the intersection points of the two given circles.
For the special values $\lambda=1;\;\mu=-1$ we get the equation of the line joining the intersection points $(a,b), (a',b')$ wich is perpendicular to the line containing all the centers of $\mathcal{C}$
$(2m'+2m)x+(2n'-2n)y+p-p'=0$