The radical axis of any two circles is dots | vedclass

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Question

(B) The radical axis of any two circles is perpendicular to the line joining their centres. Let the equations of two circles be $x^2+y^2+2g_1x+2f_1y+c

Vedclass · Accepted Answer

Perpendicular

Vedclass · Answer

(B) The radical axis of any two circles is perpendicular to the line joining their centres. Let the equations of two circles be $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$. The equation of the radical axis is obtained by subtracting the two equations: $2(g_1-g_2)x + 2(f_1-f_2)y + (c_1-c_2) = 0$ ...$(i)$ The slope of the radical axis $(i)$ is $m_1 = -\frac{g_1-g_2}{f_1-f_2}$. The centres of the circles are $C_1(-g_1, -f_1)$ and $C_2(-g_2, -f_2)$. The slope of the line joining the centres is $m_2 = \frac{-f_2 - (-f_1)}{-g_2 - (-g_1)} = \frac{f_1-f_2}{g_1-g_2}$. Since $m_1 	imes m_2 = (-\frac{g_1-g_2}{f_1-f_2}) 	imes (\frac{f_1-f_2}{g_1-g_2}) = -1$,the radical axis is perpendicular to the line joining the centres. Thus,option $B$ is correct.

The radical axis of any two circles is $ \dots $ to the line joining their centres.

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