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If the circles $(x - a)^{2} + (y - b)^{2} = c^{2}$ and $(x - b)^{2} + (y - a)^{2} = c^{2}$ touch each other, then

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a=b±2c

a=b±2 c

a=b±c

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Correct option is D

The circles will touch each other if the length of the common chord is zero i.e.4c2−2(a−b)2=0⇒2c2=(a−b)2⇒a−b=±2c

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If the circles (x−a)2+(y−b)2=c2 and (x−b)2+(y−a)2=c2 touch each other, then

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Ready to Test Your Skills?

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If the circles $(x - a)^{2} + (y - b)^{2} = c^{2}$ and $(x - b)^{2} + (y - a)^{2} = c^{2}$ touch each other, then