Answer
$x^2+y^2+6y-8x=0$
Work Step by Step
The general equation for a circle with radius $r$ and centre $(h,k)$ is: $(x-h)^2+(y-k)^2=r^2$. Hence our equation: $(x-4)^2+(y-(-3))^2=5^2\\(x-4)^2+(y+3)^2=5^2$.
$(x-4)^2+(y+3)^2=5^2$ in general form is:
$(x-4)^2+(y+3)^2=5^2\\x^2-8x+16+y^2+6y+9=25\\x^2+y^2+6y-8x=0$
