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# Find the relation so that the two circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g_1x+2f_1+c_1=0$cut each other orthogonally.

$\begin{array}{1 1}(a)\;c+c_1=2gg_1-2ff_1\\(b)\;c+c_1=2gg_1+2ff_1\\(c)\;c+c_1=gg_1-ff_1\\(d)\;c-c_1=2gg_1+2ff_1\end{array}$

Given two circles are
$x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g_1x+2f_1y+c_1=0$
Centres of the circles are $C_1(-g,-f)$ and $C_2(-g_1,f_1)$
Radius of the circles are $r_1=\sqrt{g^2+f^2-c}\:\:and\:\:r_2=\sqrt{g_1^2+f_1^2-c_1}$
If circle cut each other orthogonally then $\theta=90^{\large\circ}$
Hence $\cos(180-\theta)=\large\frac{r_1^2+r_2^2-d^2}{2r_1r_2}$
$0=r_1^2+r_2^2-d^2$
$r_1^2+r_2^2=d^2$
$g^2+f^2-c+g_1^2+f_1^2-c_1=g^2+f^2+g_1^2+f_1^2-2gg_1-2ff_1$
$c+c_1=2gg_1+2ff_1$
Hence (b) is the correct answer.
edited Mar 20, 2014