Chapter 8.6, Problem 9QR

Solution

To find: The center and radius of the circle $x^2+y^2+2x-6y+6=0$ in the $xy$ -plane.

The center is $\left(-1,3\right)$ and the radius is $2$ .

Given information:

Equation of circle: $x^2+y^2+2x-6y+6=0$

Formula used:

Standard formula of circle: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Center of the circle: $\left(h,k\right)$

Radius of the circle: $r$

Calculation:

The given equation is : $x^2+y^2+2x-6y+6=0$

Regroup by writing the $x^2$ and $x$ terms together, $y^2$ and $y$ terms together.

  $x^2+2x+y^2-6y+6=0$

Use completing the square method to bring the given equation to the form ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Add and subtract 1 to write $x^2+2x$ as a whole square and add and subtract $9$ to write $y^2-6y$ as whole square.

  $x^2+2x+1-1+y^2-6y+6+9-9=0$

The expression $x^2+2x+1$ can be written as ${\left(x+1\right)}^2$ .

The expression $y^2-6y+9$ can be written as ${\left(y-3\right)}^2$ .

  $\begin{array}{l}{\left(x+1\right)}^2+{\left(y-3\right)}^2-1-9+6=0\\ {\left(x+1\right)}^2+{\left(y-3\right)}^2-10+6=0\\ {\left(x+1\right)}^2+{\left(y-3\right)}^2-4=0\\ {\left(x+1\right)}^2+{\left(y-3\right)}^2=4\end{array}$

Therefore, the center is $\left(-1,3\right)$ and radius is $2$ .