JEE Main 2014 — Circle Question with Solution

We know that the equation of the family of circle passing through the point of intersection of a circle $S=0$ and a line $L=0$ is $S+\lambda L=0.$

Hence, the family of circle passing through points of intersection of given circle $x^2+y^2-16=0$ and chord $3x+y+5=0$ is $(x^2+y^2-16)+\lambda \left(3x+y+5\right)=0$

$\Rightarrow x^2+y^2+3\lambda x+\lambda y+5\lambda -16=0$

We know that, the centre of a circle $x^2+y^2+2gx+2fy+c=0$ is $\left(-g, -f\right)$

Hence, centre of the circle $x^2+y^2+3\lambda x+\lambda y+5\lambda -16=0$ is $\left(-\frac{3\lambda }{2}, -\frac{\lambda }{2}\right)$ and since, for this circle $AB$ is diameter, thus $\left(\frac{-3\lambda }{2}, \frac{-\lambda }{2} \right)$ must lie on $AB$ i.e. $3x+\mathit{y}+5=0$

$\Rightarrow 3\left(-\frac{3\lambda }{2}\right)+\left(-\frac{\lambda }{2}\right) +5=0$

$\Rightarrow -\frac{9\lambda }{2}-\frac{\lambda }{2}+5=0$

$\Rightarrow \lambda =1.$

Therefore, the equation of the required circle is $x^2+y^2+3\times 1\times x+1\times y+5\times 1-16=0$

$\Rightarrow x^2+y^2+3x+y-11=0.$