Q. If the circles $x^{2} + y^{2} + 2gx + 2fy = 0$ and $x^{2} + y^{2} + 2g^{'}x + 2f^{'}y = 0$ touch each other, then
A) $ff^{'} = gg^{'}$
B) $fg = f^{'}g^{'}$
C) $(fg)^{2} = \left( f^{'}g^{'} \right)^{2}$
D) $fg^{'} = f^{'}g$
Explanation
(d)
As both the circles pass through the origin and so they must have the same tangent at (0, 0). The general equation of tangent of the given circles are
$xx_{1} + yy_{1} + g\left( x + x_{1} \right) + f\left( y + y_{1} \right) = 0$
$xx_{1} + yy_{1} + g^{'}\left( x + x_{1} \right) + f^{'}\left( y + y_{1} \right) = 0$
On substituting $x_{1} = 0$ and $y_{1} = 0$, we get
$gx + fy = 0\ \ \ \Rightarrow \ \ \ g^{'}x + f^{'}y = 0$
$or\frac{f}{g} = \frac{f^{'}}{g}or{\ g}^{'}f = gf'$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.