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
$x^2+y^2-5=0,x^2+y^2-8x-6y+10=0$ 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.
$\therefore 2\left(g_1{g}_2+f_1{f}_2^{\ast }\right)=c_1+c$
$\therefore 2[-g(0)+(-f)(0)]=c-5$
$\Rightarrow c=5$
$2[-g(4)+(-f)(3)]=c+10$
$\Rightarrow -2(4g+3f)=5+10$
$\Rightarrow 4g+3f=-\frac{15}{2}...(i)$
and $2[-g(2)+(-f)(-1)]=c-2$
$\Rightarrow 2[-2g+f]=5-2$
$\Rightarrow -2g+f=\frac{3}{2}...(ii)$
On solving Eqs. (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 $4f=4\times \frac{-9}{10}$
$\Rightarrow 4f=\frac{-36}{10}$
$\Rightarrow 4f=3g$