If one of the diameters of the circle x ^ 2 + y ^ 2 - 10x + 4y + 13...

Given,

${S}_{1} ≡ {x}^{2} + {y}^{2} - 10x + 4y + 13 = 0$ with radius $R = 4 and its one diameter is chord of other circle${S}_{2}$whose centre is given by intersection of$2x + 3y = 12 \text{ and } 3x - 2y = 5$which is \left(3, 2\right)$

Now, plotting the diagram we get,

Now, from above circle the distance between the centres is given by $l = \sqrt{{\left(5 - 3\right)}^{2} + {\left(- 2 - 2\right)}^{2}} = \sqrt{20}$

Hence, radius of ${S}_{2}$ will be, ${r}^{2} = {\left(\sqrt{20}\right)}^{2} + {4}^{2} = 20 + 16 = 36$

$⇒ r = 6$