The diameters of a circle are along $2x+y-7=0$ and $x+3y-11=0$. Then the equation of the circle, which also passes through $\left(5,7\right)$ is

Given diameters are

 $2\mathrm{x}+\mathrm{y}-7=0$

 $\mathrm{x}+3\mathrm{y}-11=0$

solving above equations we get centre

centre=(2,3)

where$C=(2, 3), P=(5, 7)$

 $\mathrm{Now} \mathrm{r}=CP=\sqrt{9+16}=5$

$$\begin{gathered} So circle equation is \\ {\left(\mathrm{x}-2\right)}^2+{\left(\mathrm{y}-3\right)}^2=25 \\ \Rightarrow x^2+y^2-4x-6y+13-25=0 \\ \Rightarrow x^2+y^2-4x-6y-12=0 \end{gathered}$$