Question

The equation of the director circle of the circle $(x-2{)}^2+(y+3{)}^2=16$ is .

• A. $x^2+y^2-4x+6y=45$
• B. $x^2+y^2-4x+6y=19$
• C. $x^2+y^2-4x+6y=51$
• D. $x^2+y^2-4x+6y=77$

Answer 1

The correct option is B $x^2+y^2-4x+6y=19$

$\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{director}\mathrm{circle}\mathrm{of}\mathrm{the}\mathrm{circle}{x}^2+y^2=r^2\mathrm{is}{x}^2+y^2=2r^2.\therefore \mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{director}\mathrm{circle}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{circle}\mathrm{will}\mathrm{be}:(x-2{)}^2+(y+3{)}^2=2(4{)}^2\Rightarrow x^2+y^2-4x+6y=19$