The circles x^2 + y^2 + 2gx + 2fy = 0 and x^2 | vedclass

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(A) For two circles with centers $C_1(-g, -f)$ and $C_2(-g', -f')$ and radii $r_1 = \sqrt{g^2 + f^2}$ and $r_2 = \sqrt{g'^2 + f'^2}$ to touch external

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$f'g = g'f$

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(A) For two circles with centers $C_1(-g, -f)$ and $C_2(-g', -f')$ and radii $r_1 = \sqrt{g^2 + f^2}$ and $r_2 = \sqrt{g'^2 + f'^2}$ to touch externally,the distance between their centers must be equal to the sum of their radii.$C_1C_2 = r_1 + r_2$$\sqrt{(-g + g')^2 + (-f + f')^2} = \sqrt{g^2 + f^2} + \sqrt{g'^2 + f'^2}$Squaring both sides:$(g' - g)^2 + (f' - f)^2 = (g^2 + f^2) + (g'^2 + f'^2) + 2\sqrt{g^2 + f^2}\sqrt{g'^2 + f'^2}$$g'^2 - 2gg' + g^2 + f'^2 - 2ff' + f^2 = g^2 + f^2 + g'^2 + f'^2 + 2\sqrt{(g^2 + f^2)(g'^2 + f'^2)}$$-2(gg' + ff') = 2\sqrt{(g^2 + f^2)(g'^2 + f'^2)}$$-(gg' + ff') = \sqrt{(g^2 + f^2)(g'^2 + f'^2)}$Squaring again:$(gg' + ff')^2 = (g^2 + f^2)(g'^2 + f'^2)$$g^2g'^2 + f^2f'^2 + 2gg'ff' = g^2g'^2 + g^2f'^2 + f^2g'^2 + f^2f'^2$$2gg'ff' = g^2f'^2 + f^2g'^2$$g^2f'^2 + f^2g'^2 - 2(gf')(fg') = 0$$(gf' - fg')^2 = 0$$gf' = fg'$

The circles $x^2 + y^2 + 2gx + 2fy = 0$ and $x^2 + y^2 + 2g'x + 2f'y = 0$ touch each other externally if:

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