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If the rectangle hyperbola $(x-1)(y-2)=4$ cuts the circle $x^2+y^2+2gx+2fy+c=0$

at the points $(3,4)\;,(5,3)\;,(2,6)\;,(-1,0)$, then find the value of $g+f$ .

My try: Given that the circle $x^2+y^2+2gx+2fy+c=0$ passes through these $4$ points we have:

$9+16+6g+8f+c=0\Rightarrow 6g+8f+c=-25\cdots (1)$

$25+9+10g+6f+c=0\Rightarrow 10g+6f+c=-34\cdots (2)$

$4+36+4g+12f+c=0\Rightarrow 4g+12f+c=-40\cdots (3)$

$1+0-2g+0+c=0\Rightarrow -2g+c=-1\cdots (4)$

Now we will solve this system of equations for $g$ and $f$ .

But this is very tedious. Could someone explain to me how to solve in an easier way? Thanks.

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DXT
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    There is no royal road to enlightenment. One way or another, you end up computing the intersection of some lines. You can save yourself a little work by noting that the center of the circle is at $(-g,-f)$. – amd Sep 02 '18 at 04:26
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    Those $4$ points don't look to be concyclic. – dxiv Sep 02 '18 at 04:27
  • Thanks amd and dxiv, would you like to explain me in detail, Thanks – DXT Sep 02 '18 at 05:15
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    @DurgeshTiwari There is really nothing more to explain than my previous comment stated. The $4$ points do indeed lie on the given hyperbola, but they are not concyclic, so the so-called circle does not exist. It could be a typo in the problem, or maybe this was a trick question to begin with, and you were expected to catch that. As always, some more context could help. – dxiv Sep 02 '18 at 05:32
  • Thanks dxiv, would you like to exlain me how we can check concylicity of $4$ points, Thanks – DXT Sep 02 '18 at 05:40
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    @DurgeshTiwari if you plot the points, you'd see they're not concyclic – John Glenn Sep 02 '18 at 05:42
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    @DurgeshTiwari It's kind of obvious if you just draw a rough sketch on paper. Or, try to solve the system you came up with in the post, and you'll find that there are no solutions. Keep in mind that you have $4$ equations and $3$ unknowns, so the system is overdetermined to begin with. – dxiv Sep 02 '18 at 05:53

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