Question

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

• A. $ a=b\pm 2c $
• B. $ a=b\pm \sqrt{2}c $
• C. $ a=b\pm c $
• D. None of these

Answer 1

Answer: (B)

The circles will touch each other, if the length of the common chord is zero $ i.e.,
$ $ \therefore $ $ \sqrt{4{{c}^{2}}-2{{(a-b)}^{2}}}=0 $
$ \Rightarrow $ $ 2{{c}^{2}}={{(a-b)}^{2}} $
$ \Rightarrow $ $ a=b\pm \sqrt{2}c $