Chapter 9.3, Problem 46PPS

To determine

To Find: The center and radius of the circle then graph the circle.

Answer to Problem 46PPS

  $\begin{array}{l}(h,k)=\left(-2,1\right)\\ r=\sqrt{13}\end{array}$

Explanation of Solution

Given: $3x^2+3y^2-6y+12x=24$

Equation of the circle

  ${(x-h)}^2+{(y-k)}^2=r^2\mathrm{....}(i)$

We can write the equation $3x^2+3y^2-6y+12x=24$ as

  $\begin{array}{}$ 3x^2+3y^2-6y+12x=24$$ x^2+y^2-2y+4x=8$$ x^2+4x+4+y^2-2y+1=8+4+1$$ {(x+2)}^2+{(y-1)}^2=13$$ {(x+2)}^2+{(y-1)}^2={(\sqrt{13})}^2$\end{array}$

Comparing with $(i)$

We get

  $\begin{array}{l}(h,k)=\left(-2,1\right)\\ r=\sqrt{13}\end{array}$

Where $(h,k)$ is the center of the circle and $r$ is the radius of the circle

The graph of the circle is