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 which of he following is/are true -

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

Answer 1

The correct option is D $4f=3g$

Centres and constant terms in the circles $x^2+y^2-5=0,x^2+y^2-8x-6y+10=0$ and $x^2+y^2-4x+2y-2=0$ are $C_1^{\prime }(0,0),c_1=-5,C_2^{\prime }(4,3),c_2=10$ and $C_3^{\prime }(2,-1),c_3=-2$
Also, centre and constant of circle $x^2+y^2+2gx+2fy+c=0$ is $C_4^{\prime }(-g,-f)$ and $c_4=c$
Since, the first circle intersect all the three at the extremities of diameter, therefore they are orthogonal to each other.
$\therefore 2(g_1{g}_2+f_1{f}_2)=c_1+c$
$\therefore 2[-g\left(0\right)+(-f)\left(0\right)]=c-5$
$\Rightarrow c=5$

$2[-g\left(4\right)+(-f)\left(3\right)]=c+10$
$\Rightarrow -2(4g+3f)=5+10$
$\Rightarrow 4g+3f=-\frac{15}{2}$ .......(i)

and $2[-g\left(2\right)+(-f)(-1)]=c-2$
$\Rightarrow 2[-2g+f]=5-2$
$\Rightarrow -2g+f=\frac{3}{2}$ .......(ii)

On solving equations (i) and (ii), we get
$f=-\frac{9}{10}$ and $g=-\frac{12}{10}$
$\therefore fg=\frac{-9}{10}\times -\frac{12}{10}=\frac{27}{25}$
and $4f=4\times \frac{-9}{10}$
$\Rightarrow 4f=\frac{-36}{10}$
$\Rightarrow 4f=3g$