Show that the line 5x + 12y -4=0 touches the circle $x^2+y^2-6x+4y+12=0$

Circle x² + y²  -6x +4y+12 = 0
Its center = (3.-2) , radius = $\sqrt{9+4-12 } = 1$
the line  5x +12y -4 = 0 touches the circle if r = d
d=perpendicular distance from centre (3,-2) to the line 5x + 12 y -4 =0
   = $\frac{\left|15\left(3\right)+ 12 (-2) -41 \right|}{\sqrt[ ]{5^2 +{12}^2}} = \frac{13}{13} = 1 = \mathrm{r}$
Hence given line touches the circle.