Question

The equation of the circle which cuts the circles
$S_{1} \equiv x^{2}+y^{2}-4=0$
$S_{2} \equiv x^{2}+y^{2}-6 \,x-8 \,y+10=0$
$S_{3} \equiv x^{2}+y^{2}+2 \,x-4 \,y-2=0$
at the extremities of diameters of these circles is

• A. $x^{2}+y^{2}-4 \,x-6 \,y-4=0$
• B. $x^{2}+y^{2}+4 x-4=0$
• C. $x^{2}+y^{2}=25$
• D. $x^{2}+y^{2}+x+y+1=0$

Answer 1

Answer: (A)

Let the equation of required circle is
$S \equiv x^{2}+y^{2}+2 g x+2 f y+c=0$
and equation of given circles
$S_{1} \equiv x^{2}+y^{2}-4=0$
$S_{2} \equiv x^{2}+y^{2}-6 \,x-8\, y+10=0 $
$S_{3}=x^{2}+y^{2}+2 \,x-4,\,y-2=0$
$\because$ Circle $S=0$ cuts the circles $S_{1}=0, S_{2}=0$ and $S_{3}=0$ at the extermities of the diameters, so common chord of $S=0$ and $S_{1}=0$ passes through the centre of the circle $S_{1}=0$, so $c=-4$
Similarly $(2 \,g+6) x+(2\, f+8) y-14=0$ passes
through $(3,4)$, so $6\, g+8\, f+36=0$
$\Rightarrow 3 \,g+4 f+18=0 \ldots(i)$
and $(2 \,g-2) x+(2\, f+4) y-2=0$, passes through $(-1,2), so$
$-2 \,g+4 \,f+8=0 $ ..........(ii)
From Eqs. (i) and (ii), we get
$(g, f)=(-2,-3)$
So equation of required circle is
$x^{2}+y^{2}-4\, x-6 \,y-4=0$