Chapter 13, Problem 22RP

To determine

To find: The centre and the radius of the graph for the given equation.

Answer to Problem 22RP

The centre of the circle is $\left(-3,2\right)$ and the radius of the circle is 5.

Explanation of Solution

Given:

The given equation is $x^2+6x+y^2-4y=12$ .

Calculation:

Consider the given equation is,

  $x^2+6x+y^2-4y=12$

Solve the above equation as,

  $\begin{array}{l}{x}^2+6x+y^2-4y=12\\ x^2+6x+9+y^2-4y+4=12+9+4\\ {\left(x^2+3\right)}^2+{\left(y-2\right)}^2=25\end{array}$

The general form for the equation of the circle is,

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

Then, the centre of the circle is $\left(-3,2\right)$ and the radius of the circle is 5.