Question

If the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ cuts the three circles $x^2 + y^2 - 5 = 0, x^2 + y^2 - 8x - 6y + 10 = 0$ and $x^2 + y^2 - 4x + 2y - 2 = 0$ at the extremities of their diameters, then

• A. $C = - 5$
• B. $fg = 147/25$
• C. $g + 2f = c + 2$
• D. $4f = 3g$

Answer 1

Answer: (D)

Centres and constant terms in the circles
$x^{2}+y^{2}-5=0, x^{2}+y^{2}-8 x-6 y+10=0$ and
$x^{2}+y^{2}-4 x+2 y-2=0$ are $C_{1}^{'}(0,0), c_{1}=-5$
$C_{2}^{'}(4,3), c_{2}=10$ and $C_{3}^{'}(2,-1), c_{3}=-2$
Also, centre and constant of circle
$x^{2}+y^{2}+2 g x+2 f y+c=0$
is$C_{4}^{'}(-g,-f) $ and $ c_{4}=c$
Since, the first circle intersect all the three at the extremities of diameter, therefore they are orthogonal to each other.
$\therefore 2\left(g_{1}\, g_{2}+f_{1} f_{2}^{*}\right) =c_{1}+c$
$\therefore 2[-g(0)+(-f)(0)] =c-5$
$\Rightarrow c=5$
$ 2[-g(4)+(-f)(3)]=c+10 $
$ \Rightarrow -2(4 g+3 f)=5+10 $
$ \Rightarrow 4 g+3 f=-\frac{15}{2} \,\,\,...(i)$
and $2[-g(2)+(-f)(-1)]=c-2 $
$ \Rightarrow 2[-2 g+f]=5-2 $
$ \Rightarrow -2 g+f=\frac{3}{2} \,\,\,...(ii)$
On solving Eqs. (i) and (ii), we get
$f=-\frac{9}{10}$ and $g=-\frac{12}{10}$
$\therefore f g=\frac{-9}{10} \times-\frac{12}{10}=\frac{27}{25}$
and $4 f=4 \times \frac{-9}{10}$
$\Rightarrow 4 f=\frac{-36}{10}$
$\Rightarrow 4 f=3 g$