Question

What is the standard form of the equation of the circle x2 -4x + y2 + 6y + 12 = 0?

Answer 1

The standard form of the equation of the circle:


`(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-a\ coordinates\ of\ the\ circle\\r-a\ radius`


`x^2-4x+y^2+6y+12=0\\\\x^2-2x\cdot2+y^2+2y\cdot3=-12\\\\x^2-2x\cdot2+2^2+y^2+2y\cdot3+3^2=-12+2^2+3^2\ \ \ |use\ (a\pm b)^2=a^2\pm2ab+b^2\\\\(x-2)^2+(y+3)^2=-12+4+9\\\\\boxed{(x-2)^2+(y+3)^2=1}`