Question

If the two circles $x^2+y^2+2gx+2fy=0$ and $x^2+y^2+2g^{\prime }x+2f^{\prime }y=0$ touch each other, then show that $f^{\prime }g=f{g}^{\prime }$.

Answer 1

Given that:

$C_1:x^2+y^2+2gx+2fy=0$

$C_2:x^2+y^2+2g^{\prime }x+2f^{\prime }y=0$

To show:

$f^{\prime }g=f{g}^{\prime }$

Solution:

Let $C_1$ and $C_2$ be the centres of the two circles and ${}^{\prime }{O}^{\prime }$ is the origin.

Both the circles passes through origin ${}^{\prime }{O}^{\prime }$ because constant term of both the equations of the circles are $0.$

So; slope of $O{C}_1=$ slope of $O{C}_2$

or, $\frac{f-0}{g-0}=\frac{f^{\prime }-0}{g^{\prime }-0}$

or, $\frac{f}{g}=\frac{f^{\prime }}{g^{\prime }}$

or, $f{g}^{\prime }=f^{\prime }g$

or, $f^{\prime }g=f{g}^{\prime }$