Answer:
[tex]x^{2}+y^{2}-8x-4y+11=0[/tex]
Step-by-step explanation:
we have
[tex](x-4)^{2}+(y-2)^{2}=9[/tex]
The general equation of the circle is equal to
[tex]x^{2}+y^{2}+Ax+By+C=0[/tex]
Convert the standard form to a general form
[tex](x-4)^{2}+(y-2)^{2}=9\\ \\(x^{2} -8x+16)+(y^{2}-4y+4)=9\\ \\x^{2}+y^{2}-8x-4y+20-9=0\\ \\x^{2}+y^{2}-8x-4y+11=0[/tex]
