Given,

The perimeter of a triangle is $20$ and two of its vertices are $\left(-5,0\right)$ and $\left(6,0\right)$,

Now let the third vertex be $\left(x,y\right)$,

Now by formula of perimeter of triangle we get,

$20=\sqrt{{\left(x+5\right)}^2+y^2}+\sqrt{{\left(6+5\right)}^2}+\sqrt{{\left(x-6\right)}^2+y^2}$

$\Rightarrow 20=\sqrt{{\left(x+5\right)}^2+y^2}+11+\sqrt{{\left(x-6\right)}^2+y^2}$

$\Rightarrow 9-\sqrt{{\left(x-6\right)}^2+y^2}=\sqrt{{\left(x+5\right)}^2+y^2}$

Now squaring both side we get,

$\Rightarrow 9^2+{\left(x-6\right)}^2+y^2-18\sqrt{{\left(x-6\right)}^2+y^2}={\left(x+5\right)}^2+y^2$

$\Rightarrow 9^2+{\left(x-6\right)}^2-{\left(x+5\right)}^2=18\sqrt{{\left(x-6\right)}^2+y^2}$

$\Rightarrow 92-22x=18\sqrt{{\left(x-6\right)}^2+y^2}$

$\Rightarrow {\left(46-11x\right)}^2=9^2\left({\left(x-6\right)}^2+y^2\right)$

$\Rightarrow 40x^2-81y^2-40x-800=0$