• question_answer The equation of the circle which cuts each of the three circles ${{x}^{2}}+{{y}^{2}}=4,$${{(x-1)}^{2}}+{{y}^{2}}=4$ and ${{x}^{2}}+{{(y-2)}^{2}}=4$ orthogonally is A)  ${{x}^{2}}+{{y}^{2}}+x+2y+4=0$ B)  ${{x}^{2}}+{{y}^{2}}+x-2y+4=0$ C)  ${{x}^{2}}+{{y}^{2}}-x-2y+4=0$ D)  ${{x}^{2}}+{{y}^{2}}-x+2y+4=0$

Given,             $\Rightarrow$ $[\because |adj\,(adj\,A)|=\,A{{|}^{{{(n-1)}^{2}}}}]$ and                  $=100\,\left( \frac{\Delta \phi }{\Delta t} \right)$ Let the required circle be $=5\times {{10}^{-3}}\times 100=0.5\,\,V$ Now, by condition of orthoganality,$R=v\sqrt{\frac{2h}{g}}$         ?(i) $u=v,$            $u=v+at$                  ?(ii) and      $0=v-at$                   ?(iii) $\therefore$    $-a=\frac{0-v}{t}=-\frac{v}{t}$      and      $f=\mu R=\mu mg$ $a=\mu g$      $\therefore$    [from Eq. (i)] $t=\frac{v}{a}=\frac{v}{g\mu }$    $\eta =\frac{{{P}_{0}}}{{{P}_{i}}}$       $\therefore$    Required equation of circle $\frac{1}{2}\,m{{v}^{2}}=16\,\,J$ $v=4\,m{{s}^{-1}}$