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 ?





Can't think of a simple method.Help please!

  • $\begingroup$ This may not be easiest but. As you have the equations you can find the points of intersection of the first circle and the second circle. As these two points will be opposite sides of the diameter you know how far apart the points are. From that you can solve for g and f and c. $\endgroup$ – fleablood Jul 16 '16 at 6:05
  • $\begingroup$ Not a very good idea.It was asked in a competitive exam with 2-3 mins per question.... @fleablood $\endgroup$ – user220382 Jul 16 '16 at 6:06
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    $\begingroup$ What does "at the extremities of their diameters" mean? $\endgroup$ – mvw Jul 16 '16 at 6:16
  • $\begingroup$ I assume it means the points are endpoints of a diameter. I.e. they are 2r apart. Big circle has center -g, -f and radius f^2+g^2-v^2 First circle has center 0,0 and radius root 5. 2nd circle center 4,3 and radius root 15. 3rd circle center 2,-1 and radius root 7. Big circle and first circ intersect 2gx +2fy + c = -2gx -2fy +c =-5 and x^2 + y^2 = 5. From that c = -5. And g =-f so that is the only one. $\endgroup$ – fleablood Jul 16 '16 at 22:00
  • $\begingroup$ So you can do my answer in under 2 minutes. I didn't but someone could. $\endgroup$ – fleablood Jul 16 '16 at 22:00

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 $f=-21/10, g=-14/5$.

  • $\begingroup$ Very very intelligent method.Great :-)! $\endgroup$ – user220382 Jul 16 '16 at 6:30

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