9 Distance, Midpoint and \( 1 \leftarrow \quad \)\begin{tabular}{l} Complete the square and write the given equation in standard form. Then give the center and \\ radius of the circle and graph the equation. \\ \hline\end{tabular} \( x^{2}+y^{2}+4 x+6 y+12=0 \)

To rewrite the given equation in standard form, we need to complete the square for both the \(x\) and \(y\) terms. Starting with the original equation: \[ x^{2} + y^{2} + 4x + 6y + 12 = 0 \] We can rearrange it: \[ x^{2} + 4x + y^{2} + 6y = -12 \] Now, completing the square for \(x\): Take the coefficient of \(x\), which is \(4\), divide by \(2\) to get \(2\), and square it to get \(4\). We then rewrite: \[ (x^2 + 4x + 4) \rightarrow (x + 2)^{2} \] Next, for \(y\): Take the coefficient of \(y\), which is \(6\), divide by \(2\) to get \(3\), and square it to get \(9\). We then rewrite: \[ (y^2 + 6y + 9) \rightarrow (y + 3)^{2} \] Now, we add these squares to the other side of the equation: \[ (x + 2)^{2} + (y + 3)^{2} = -12 + 4 + 9\] This simplifies to: \[ (x + 2)^{2} + (y + 3)^{2} = 1 \] Now we have the equation in standard form, which represents a circle. The center of the circle is at \((-2, -3)\) and the radius is \(\sqrt{1} = 1\). To graph the equation, plot the center at \((-2, -3)\) and draw a circle with a radius of \(1\), which will touch the points at \((-3, -3)\), \((-1, -3)\), \(-2, -2)\), and \((-2, -4)\). Happy graphing!