We have the equation:
[tex]x^2+y^2-8x-6y+24=0[/tex]
By arranging this equation in terms of x and y, we have:
[tex]x^2-8x+y^2-6y=-24 \\ \\[/tex]
By using the method of completing the square, we have:
[tex]x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}[/tex]
The center of this circle is:
[tex](h,k)=(4,3)[/tex]
So the equation that fulfills the statement is:
[tex](x-4)^2+(y-3)^2=2^2[/tex]
Finally, the right answer is c)
