If two circles $a (x^{2} + y^{2}) + b x + c y = 0$ and $A (x^{2} + y^{2}) + B x + C y = 0$ touch each other, then

a c = c A

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b C = c B

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a B = b A

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a A = b B = c C

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Solution

The correct option is B $b C = c B$
Given equation of circles are $a (x^{2} + y^{2}) + b x + c y = 0$ and $A (x^{2} + y^{2}) + B x + C y = 0$
Clearly, both the circles pass through the origin.
The two circle will touch each other, if they have a common tangents at the origin. The tangents at the origin.
The tangents at the origin to the two circles are $b x + c y = 0$ and $B x + C y = 0$ . These two must be identical
Therefore, $\frac{b}{B} = \frac{c}{C} \Rightarrow b C = c B$

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