If $4a^2 – 5b^2 + 6a + 1 = 0$  and the line $ax + by + 1 = 0$ touches a fixed circle, then

$4a^2 – 5b^2 + 6a + 1 = 0$

$9{\mathrm{a}}^2+6\mathrm{a}+1=5\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)$

$\frac{3\mathrm{a}+1}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=\sqrt{5}$
$\Rightarrow {\perp }_{\mathrm{ar}}$distance of (3, 0) from  to the line
$\mathrm{ax}+\mathrm{by}+1\equiv 0 \mathrm{is} \sqrt{5}$

Equation of circle is ${\left(x-3\right)}^2+y^2=5$