If the circle $x^2+y^2+2gx+2fy+c=0$ cuts the three circles $x^2+y^2−5=0$, $x^2+y^2−8x−6y+10=0$ and $x^2+y^2−4x+2y−2=0$ at the extremities of their diameters, then which of the following are true ?
$c=-5$
$fg=147/25$
$g+2f=c+2$
$4f=3g$
Can't think of a simple method.Help please!
The center of $x^2+y^2−5=0, x^2+y^2−8x−6y+10=0$ and $x^2+y^2−4x+2y−2=0$ is $(0,0), (4,3)$ and $(2,-1)$ respectively.
So, $$(x^2+y^2+2gx+2fy+c)-(x^2+y^2-5)=0$$ $$(x^2+y^2+2gx+2fy+c)-(x^2+y^2−8x−6y+10)=0$$ $$(x^2+y^2+2gx+2fy+c)-(x^2+y^2−4x+2y−2)=0$$ passes through $(0,0), (4,3)$ and $(2,-1)$ respectively.
(For example, $(x^2+y^2+2gx+2fy+c)-(x^2+y^2-5)=0$ represents a line passing through the intersection points of the two circles.)
Therefore, we get $$c=-5$$ $$c+6 f+8 g+40 = 0$$ $$c-2 f+4 g+12 = 0$$ Solving the system gives $f=-21/10, g=-14/5$.