Question

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

• A. $(2,3)$
• B. $(-2,3)$
• C. $(2,-3)$
• D. $(-2,-3)$

Answer 1

The correct option is A $(2,3)$

$s:x^2+y^2+2gx+2fy+c=0$
$s_1:x^2+y^2=4$ centre$=(0,0)$
$s_2:x^2+y^2=6x-8t+10=0$ centre $=(3,4)$
$s_3:x^2+y^2+2x-4y-2=0$ centre$=(-1,2)$
So, radial axis of $s$ and $s_1$ passes through $(0,0)$
$2gx+2fy+c+4=0$
$c=-4$
So, radical axis of $s$ and $s_2$ passes through $(Q,1).$
$(2g+6)x+(28+8)y-14=0$
$6g+18+8f+18=0$
$3g+4f+18=0--\left(1\right)$
So, radical axis of $s$ and $s_3$ pass through $(-1,2)$
$(29-2)x+(2f+4)y-2=0$
$-2g+2+4f+6=0$
$2f-g+4=0--\left(1\right)$
from (1) & (2)
$10f+30=0$
$f=-3$ and $g=-2$
So, centre of circle is $(-g,-f)=(2,3)$