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 $i.e.,<br/>$ $\therefore$ $\sqrt{4c^2-2{(a-b)}^2}=0$
$\Rightarrow$ $2c^2={(a-b)}^2$
$\Rightarrow$ $a=b\pm \sqrt{2}c$