If the two circles,x^2 + y^2 + 2g_1x + 2f_1y | vedclass

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(B) Both circles pass through the origin $(0, 0)$.Let the centers be $C_1 = (-g_1, -f_1)$ and $C_2 = (-g_2, -f_2)$.Since the circles touch each other

Vedclass · Accepted Answer

$\frac{f_1}{g_1} = \frac{f_2}{g_2}$

Vedclass · Answer

(B) Both circles pass through the origin $(0, 0)$.Let the centers be $C_1 = (-g_1, -f_1)$ and $C_2 = (-g_2, -f_2)$.Since the circles touch each other at the origin $(0, 0)$,the centers $C_1$,$C_2$,and the point of contact $(0, 0)$ must be collinear.The slope of the line joining $C_1$ and $(0, 0)$ is $\frac{-f_1 - 0}{-g_1 - 0} = \frac{f_1}{g_1}$.The slope of the line joining $C_2$ and $(0, 0)$ is $\frac{-f_2 - 0}{-g_2 - 0} = \frac{f_2}{g_2}$.Since the points are collinear,the slopes must be equal:$\frac{f_1}{g_1} = \frac{f_2}{g_2}$.

If the two circles,$x^2 + y^2 + 2g_1x + 2f_1y = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y = 0$ touch each other,then:

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