Question

If the circle $x^2+y^2+2gx+2fy+c=0$ cuts each of the circles $x^2+y^2-4=0,x^2+y^2-6x-8y+10=0$ and $x^2+y^2+2x-4y-2=0$ at the extremities of a diameter, then

• A. $c=-4$
• B. $g+f=c-1$
• C. $g^2+f^2-c=17$
• D. $gf=6$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $c=-4$
B $g+f=c-1$
C $g^2+f^2-c=17$
D $gf=6$

Since the circle $x^2+y^2+2gx+2fy+c=0$ cuts the three given circles at the extremities of a diameter, the common chords will pass through the centre of the respective circles, so that
$2gx+2fy+c+4=0$ passes through $(0,0)$ $\Rightarrow c=-4\left(1\right)$
Next $2gx+2fy+c+6x+8y-10=0$ passess through $(3,4)$ $\Rightarrow (2g+6)3+(2f+8)4-14=0$ $\Rightarrow 3g+4f+18=0\left(ii\right)$ and $2gx+2fy+c-2x+4y+2=0$ passes through $(-1,2)$
$\Rightarrow (2g-2)(-1)+(2f+4)2-2=0$ $\Rightarrow g-2f-4=0\left(iii\right)$
From $\left(ii\right)$ and $\left(iii\right)$ we get $g=-2$ and $f=-3.$
$\Rightarrow g+f=c-1=-5$
$g^2+f^2-c=4+9+4=17$
$gf=6,$
Ans: A,B,C,D