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If the circles x2 + y2 + 2ax + c = 0 and x2 + y2 + 2by + c = 0 touch each other,

A. \(\cfrac1{a^2}+\cfrac1{b^2}=\cfrac 1c\)

B. \(\cfrac1{a^2}+\cfrac1{b^2}=\cfrac 1{c^2}\)

C. a + b = 2c

D. \(\cfrac1a+\cfrac1b=\cfrac2c\)

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Correct option is  A. \(\cfrac1{a^2}+\cfrac1{b^2}=\cfrac 1c\) 

Given that the circles x2 + y2 + 2ax + c = 0 and x2 + y2 + 2by + c = 0 touch each other externally (assumed).

We need to find the relation of a, b and c.

We know that if the circles touch each other externally, the distance between the centres is equal to the sum of the radii of two circles.

We know that for a circle x2 + y2 + 2ax + 2by + c = 0

⇒ Centre = (- a, - b)

⇒ Radius = \(\sqrt{a^2+b^2-c}\)

For x2 + y2 + 2ax + c = 0

⇒ C1 = (- a, 0)

⇒ Radius(r1) = \(\sqrt{(-a)^2+0-c}\)

⇒ r1 = \(\sqrt{a^2-c}\)

For x2 + y2 + 2by + c = 0

⇒ C2 = (0, - b)

⇒ Radius(r2) = \(\sqrt{0^2+(-b)^2-c}\)

⇒ r2 = \(\sqrt{b^2-c}\) 

We have C1C2 = r1 + r2

⇒ c2 = a2b2 - a2c - b2c + c2

⇒ a2b2 = a2c + b2c

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