If the circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0$ and $x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y = 0$ touch each other prove that:
$f_{1} g_{2} = f_{2} g_{1}$

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Solution

The equation of the circle after completing squares can be written as
$x^{2} + 2 g_{1} x + g_{1}^{2} + y^{2} + 2 f_{1} y + f_{1}^{2} = g_{1}^{2} + f_{1}^{2} ⟹ (x + g_{1})^{2} + (y + f_{1})^{2} = g_{1}^{2} + f_{1}^{2}$
$x^{2} + 2 g_{2} x + g_{2}^{2} + y^{2} + 2 f_{2} y + f_{2}^{2} = g_{2}^{2} + f_{2}^{2} ⟹ (x + g_{2})^{2} + (y + f_{2})^{2} = g_{2}^{2} + f_{2}^{2}$
Now, as these are touching, the distance between centres is equal to sum of the radii
$\sqrt{(- g_{1} + g_{2})^{2} + (- f_{1} + f_{2})^{2}} = \sqrt{g_{1}^{2} + f_{1}^{2}} + \sqrt{g_{2}^{2} + f_{2}^{2}}$
Squaring both sides,
$(- g_{1} + g_{2})^{2} + (- f_{1} + f_{2})^{2} = g_{1}^{2} + f_{1}^{2} + g_{2}^{2} + f_{2}^{2} + 2 \sqrt{(g_{1}^{2} + f_{1}^{2}) (g_{2}^{2} + f_{2}^{2})}$
$∴ g_{1}^{2} + g_{2}^{2} - 2 g_{1} g_{2} + f_{1}^{2} + f_{2}^{2} - 2 f_{1} f_{2} = g_{1}^{2} + f_{1}^{2} + g_{2}^{2} + f_{2}^{2} + 2 \sqrt{(g_{1}^{2} + f_{1}^{2}) (g_{2}^{2} + f_{2}^{2})}$
$∴ - (g_{1} g_{2} + f_{1} f_{2}) = \sqrt{(g_{1}^{2} + f_{1}^{2}) (g_{2}^{2} + f_{2}^{2})}$
squaring both sides,
$(g_{1} g_{2})^{2} + (f_{1} f_{2})^{2} + 2 g_{1} g_{2} f_{1} f_{2} = (g_{1} g_{2})^{2} + (f_{1} f_{2})^{2} + (f_{1} g_{2})^{2} + (g_{1} f_{2})^{2}$
$∴ 2 g_{1} g_{2} f_{1} f_{2} = (f_{1} g_{2})^{2} + (g_{1} f_{2})^{2}$
$∴ (f_{1} g_{2})^{2} + (g_{1} f_{2})^{2} - 2 g_{1} g_{2} f_{1} f_{2} = 0$
$∴ (f_{1} g_{2} - g_{1} f_{2})^{2} = 0$
$∴ f_{1} g_{2} = g_{1} f_{2}$
Hence proved.

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If the circles x2+y2+2g1x+2f1y=0 and x2+y2+2g2x+2f2y=0 touch each other prove that:f1g2=f2g1

If the circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0$ and $x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y = 0$ touch each other prove that:
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