Question

part a: complete the square to rewrite the following equation in standard form. show all necessary work. (6 points)
x² + 4x + y² - 6y = -4
part b: what are the center and radius of the circle? (4 points)

Explanation:

Step1: Complete the square for x - terms

For the $x^{2}+4x$ part, we take half of the coefficient of $x$ (which is $4$), square it. Half of $4$ is $2$, and $2^{2}=4$. We add $4$ to both sides of the equation:
$x^{2}+4x + 4+y^{2}-6y=-4 + 4$
$(x + 2)^{2}+y^{2}-6y=0$

Step2: Complete the square for y - terms

For the $y^{2}-6y$ part, half of the coefficient of $y$ (which is $-6$) is $-3$, and $(-3)^{2}=9$. We add $9$ to both sides of the equation:
$(x + 2)^{2}+y^{2}-6y+9=0 + 9$
$(x + 2)^{2}+(y - 3)^{2}=9$

Step3: Identify the center and radius

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center of the circle and $r$ is the radius.
Comparing $(x + 2)^{2}+(y - 3)^{2}=9$ with the standard form, we have $a=-2$, $b = 3$ and $r^{2}=9$, so $r = 3$.

Answer:

Part A: $(x + 2)^{2}+(y - 3)^{2}=9$
Part B: Center: $(-2,3)$; Radius: $3$