The equation of the circle described on the chord 3x + y + 5 = 0 of...

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 $S + \lambdaL = 0 .$

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 $( {x}^{2} + {y}^{2} - 16 ) + \lambda\left(3x + y + 5\right) = 0$

$⇒ {x}^{2} + {y}^{2} + 3\lambdax + \lambday + 5\lambda - 16 = 0$

We know that, the centre of a circle ${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$ is \left(- g, - f\right)$</p><p>Hence, centre of the circle${x}^{2} + {y}^{2} + 3\lambdax + \lambday + 5\lambda - 16 = 0$is \left(- \dfrac{3\lambda}{2}, - \dfrac{\lambda}{2}\right)$ and since, for this circle $AB$ is diameter, thus $\left(\dfrac{- 3\lambda}{2}, \dfrac{- \lambda}{2}\right)$ must lie on $AB$ i.e. $3x + y + 5 = 0$

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

$⇒ - \dfrac{9\lambda}{2} - \dfrac{\lambda}{2} + 5 = 0$

$⇒ \lambda = 1 .$

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

$⇒ {x}^{2} + {y}^{2} + 3x + y - 11 = 0 .$