The radical axis of any two circles is to the line joining their centres

The radical axis of any two circles is perpendicular to the line joining their centres. Let, the equation of two circles are \(\begin{aligned} x^2+y^2+2 g_1 x+2 f_1 y+c_1 & =0 \\ x^2+y^2+2 g_2 x+2 f_2 y+c_2 & =0 \end{aligned}\) and \(x^2+y^2+2 g_2 x+2 f_2 y+c_2=0\) The equation of the radical axis is \(2\left(g_1-g_2\right) x+2\left(f_1-f_2\right) y+\left(c_1-c_2\right)=0\) ...(i) \(\because\) Slope of line (i) is \(-\frac{g_1-g_2}{f_1-f_2}=m_1\) (let)...(ii) and slope line joining centres of the circles is \(\begin{aligned} \frac{f_1-f_2}{g_1-g_2} & =m_2 \quad \ldots \text{(let)} \\ \because \quad m_1 m_2 & =-\frac{g_1-g_2}{f_1-f_2} \times \frac{f_1-f_2}{g_1-g_2}=-1 \end{aligned}\) Hence, option (b) is correct.