• # question_answer The equation of circle described on the chord $3x+y+5=0$of the circle ${{x}^{2}}+{{y}^{2}}=16$as diameter is:     JEE Main Online Paper (Held On 19 April 2016) A) ${{x}^{2}}+{{y}^{2}}+3x+y-11=0$ B) ${{x}^{2}}+{{y}^{2}}+3x+y+1=0$ C) ${{x}^{2}}+{{y}^{2}}+3x+y-2=0$ D) ${{x}^{2}}+{{y}^{2}}+3x+y-22=0$

Given circle is ${{x}^{2}}+{{y}^{2}}-16=0$ Eqn of chord say AB of given circle is $3x+y+5=0.$ Equation of required circle is ${{x}^{2}}+{{y}^{2}}-16+\lambda (3x+y+5)=0$ $\Rightarrow$${{x}^{2}}+{{y}^{2}}+(3\lambda )x+(\lambda )y+5\lambda -16=0$?(1) Centre $C=\left( \frac{-3\lambda }{2},\frac{-\lambda }{2} \right).$ If line AB is the diameter of circle (1), then $C=\left( \frac{-3\lambda }{2},\frac{-\lambda }{2} \right)$will lie on line AB. i.e. $3\left( \frac{-3\lambda }{2} \right)+\left( \frac{-\lambda }{2} \right)+5=0$ $\Rightarrow$$\frac{9\lambda -\lambda }{2}+5=0\Rightarrow \lambda =1$ Hence, required eqn of circle is ${{x}^{2}}+{{y}^{2}}+3x+y+5-16=0$ $\Rightarrow$${{x}^{2}}+{{y}^{2}}+3x+y-11=0$