Question

Solve circle equations to standard equations help me now sir

Answer 1

The given equation is

`x^2+y^2-6x+4y+12=0`

First, put in order the terms with the same variable, and the constant term goes to the other side of the equation.

`x^2-6x+y^2+4y=-12`

Second, find the constant term for each trinomial (x and y). To do so, we take each linear term and divide them by 2, to then elevate it to the square power.

`\begin{gathered} ((6)/(2))^2=3^2=9 \\ ((4)/(2))^2=2^2=4 \end{gathered}`

Third, add the new terms on both sides of the equation.

`x^2-6x+9+y^2+4y+4=-12+9+4`

Fourth, form the square binomials.

`(x-3)^2+(y-2)^2=1`

Once we have the circle equation in standard form, we can deduct the center and the radius. In this case, the center is (3,2) and the radius is 1.

Therefore, the circle equation in standard form is

`(x-3)^2+(y-2)^2=1`