Question

Show that the equation represents a circle, and find the center and radius of the circle.

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

Short answer

The Centre of the circle is$(2,3)$

The radius of circle is 1.

Step-by-step solution

Step 1. Determining general Equation of circle.

An equation of the circle with center$(h,k)$ and radius r is

${(x-h)}^2+{(y-k)}^2=r^2$

This is called the standard form for the equation of the circle.

Step 2. Representing the equation is a circle. 

$\begin{array}{c}{x}^2+y^2+4x-6y+12=0\\ (x^2+4x)+(y^2-6y)=-12\left(\mathrm{Groupterms}\right)\\ (x^2+4x+4)+(y^2-6y+9)=-12+4+9(\mathrm{Complete}\mathrm{the}\mathrm{square}\mathrm{by}\mathrm{adding}4\mathrm{and}9\mathrm{to}\mathrm{each}\mathrm{side})\\ or{\left(x+2\right)}^2+{\left(y-3\right)}^2=1(\mathrm{Factor}\mathrm{and}\mathrm{simplify})\end{array}$

Step 3. Determining the radius and center of the circle.

${(x-2)}^2+{(y-3)}^2=1$

Comparing the equation with the standard equation of a circle

$h=2,k=3,r=1$

So, the Centre of the circle is$(2,3)$ and the radius of the circle is