The circle passing through the intersection of the circles, x ^ 2 +...

By family of circle, circle passing through intersection of given circles will be member of

${S}_{1} + {\lambda S}_{2} = 0\text{family} ( \lambda \ne 1 )$

\left({x}^{2} + {y}^{2} - 6x\right) + \lambda\left({x}^{2} + {y}^{2} - 4y\right) = 0$</p><p>$( \lambda + 1 ) {x}^{2} + ( \lambda + 1 ) {y}^{2} - 6x - 4\lambda y = 0$</p><p>${x}^{2} + {y}^{2} - \dfrac{6}{\lambda + 1}x - \dfrac{4\lambda}{\lambda + 1}y = 0$</p><p>$\text{Centre}\left(\dfrac{3}{\lambda + 1}, \dfrac{2\lambda}{\lambda + 1}\right)$</p><p>Centre lies on$2x - 3y + 12 = 0$</p><p>$2\left(\dfrac{3}{\lambda + 1}\right) - 3\left(\dfrac{2\lambda}{\lambda + 1}\right) + 12 = 0$</p><p>$6\lambda + 18 = 0$</p><p>$\lambda = - 3$</p><p>Now circle is:${x}^{2} + {y}^{2} + 3x - 6y = 0

Now check each option