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If the circle

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

cuts the three circles

$$x^2 + y^2 – 5 = 0\space;\space x^2 + y^2 – 8x – 6y + 10 = 0 \space;\space x^2 + y^2 – 4x + 2y – 2 = 0;$$ at the extremities of their diameters, then

(A) $c = – 5 $

(B) $fg = 147/25$

(C) $g + 2f = c + 2 $

(D) $4f = 3g$

I'm not understanding a single thing. Please Help

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  • $\begingroup$ What part of it do you not understand? $\endgroup$ – Obinna Nwakwue May 1 '16 at 21:47
  • $\begingroup$ Start by sketching the circles and locate the 3 points. $\endgroup$ – Narasimham May 1 '16 at 21:58
  • $\begingroup$ In equation (A), is it a $J$ or a $g$ ? Because I don't see any $J$ ? $\endgroup$ – Jean Marie May 1 '16 at 22:05
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    $\begingroup$ Interesting question. The extremities of a varying diameter of a circle vary. In addition, we have three of those. (1) I don’t think drawing a picture can locate the correct points; (2) Could it be that some constraints were missing in the question? $\endgroup$ – Mick May 2 '16 at 5:56
  • $\begingroup$ @Mick I think that the circle in question passes through both the extremities of a single diameter of any of the given fixed circles. $\endgroup$ – najayaz May 10 '16 at 6:50
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You might know that the equation of the common chord of two circle whose standard equations are $S_1=0$ and $S_2=0$ , is $S_1=S_2$ .

The required circle bisects the circumference of all the other 3 circles, so its common chord with these circles should pass through their respective centres.

So, we have: $$c=-5\tag{i}$$

$$8g+6f+c=-32-18+10\tag{ii}$$

$$4g-2f+c=-8-2-2\tag{iii}$$

On solving above system of equations you can get the values of $c,f,g$.

As requested by Mick, here is a picture :

enter image description here

I've marked the centres and diameters of the given circles, and the required circle is shown in red.

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  • $\begingroup$ After reversing everything you did, I have the picture drawn and have the question solved your way. $\endgroup$ – Mick May 10 '16 at 10:30
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    $\begingroup$ Good work. I am also planning on uploading mine. $\endgroup$ – Mick May 10 '16 at 10:55

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