Answer : $(x-3)^2+(y+4)^2=\big(\large\frac{45}{13}\big)^2$
Given the coordinates of the centre is $(3,-4)$ and the equation of the line touching the circle is $5x+12y-12=0$
Hence $r=\bigg|\large\frac{5(3)+12(-4)-12}{\sqrt{25+144}}\bigg|$
$\Rightarrow r= \bigg|\large\frac{15-48-12}{13}\bigg|$
$\Rightarrow r= \large\frac{45}{13}$
Hence the equation of the circle is $(x-3)^2+(y+4)^2=\big(\large\frac{45}{13}\big)^2$