• # question_answer The value of k for which the circles ${{x}^{2}}+{{y}^{2}}-~3x+ky-5=0$and $4{{x}^{2}}+4{{y}^{2}}-12x-y-9=0$became concentric is A)  $-\frac{1}{6}$                         B)  $\frac{1}{6}$              C)  $-\frac{1}{4}$                         D)  $\frac{1}{4}$

[c]  ${{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}-3x+ky-5=0.$ ${{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}-3x-\frac{y}{4}-\frac{9}{4}=0.$ ${{S}_{1}}\text{ }\!\!\And\!\!\text{ }{{S}_{2}}$be concentric i.e. they have the common centre Centre of ${{S}_{1}}\equiv \left( \frac{3}{2},\frac{-k}{2} \right),$Centre of ${{S}_{2}}\equiv \left( \frac{3}{2},\frac{1}{8} \right),$ $\Rightarrow \frac{-k}{2}=\frac{1}{8}\Rightarrow k=\frac{-1}{4}$ Hence, option [c] is correct.