Answer:
[tex]\large\boxed{x^2+y^2-10x+6y-47=0}[/tex]
Step-by-step explanation:
The equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
We have the center (5, -3) and the radius r = 9.
Substitute:
[tex](x-5)^2+(y-(-3))^2=9^2\\\\(x-5)^2+(y+3)^2=81[/tex]
Use
[tex](a\pm b)^2=a^2\pm2ab+b^2[/tex]
[tex]x^2-2(x)(5)+5^2+y^2+2(y)(3)+3^2=81\\\\x^2-10x+25+y^2+6y+9=81\\\\x^2+y^2-10x+6y+(25+9)=81\\\\x^2+y^2-10x+6y+34=81\qquad\text{subtract 81 from both sides}\\\\x^2+y^2-10x+6y-47=0[/tex]
