If the two circles x 2- y 2-2 g x -2 f y -0 and x 2- y 2-2 g 1 x -2 f 1 y -0 touch each other thenA- ff 1- gg 1B- f g1-f1 g1C- fg- 1- f - 1 g -D- f 2- g 2- f 12- f 12

The correct option is C (fg_1 = f_1g |r_1 ± r_2|=C_1 C_2 |√(g^2+f^2)±√(g_1^2+f_1^2)| = √((g g_1)^2+(f f_1)^2) ⇒ gf_1 = fg_1 ...

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Coaxial System

If the two ci...

Question

If the two circles $x^{2} + y^{2} + 2 g x + 2 f y = 0$ and $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0$ touch each other then

$f f_{1} = g g_{1}$

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(fg_1 = f_1g\)

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$f^{2} + g^{2} = f_{1}^{2} + f_{1}^{2}$

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$f g_{1} = f_{1} g_{1}$

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If the two circles $x^{2} + y^{3} + 2 g x + 2 f y = 0 a n d x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0$ touch each other then

x^{2} + y^{2} + 2 g x + 2 f y = 0

x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0

x^{2} + y^{2} + 2 g x + 2 f y = 0

x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0

x^{2} + y^{2} + 2 g x + 2 f y = 0

x^{2} + y^{2} + 2 g^{1} x + 2 f^{1} y = 0

x^{2} + y^{2} + 2 g x + 2 f y = 0

x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0

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