Question

using the following equation, find the center and radius: x² + 4x + y² - 6y = -4. the center is located at (-2, 3), and the radius is 3. the center is located at (2, -3), and the radius is 3. the center is located at (-2, 3), and the radius is 9. the center is located at (2, -3), and the radius is 9.

Explanation:

Step1: Complete the square for x - terms

For the $x^{2}+4x$ part, we know that $(x + a)^{2}=x^{2}+2ax + a^{2}$. Here $2a = 4$, so $a = 2$ and $x^{2}+4x=(x + 2)^{2}-4$.

Step2: Complete the square for y - terms

For the $y^{2}-6y$ part, since $2a=-6$, then $a=-3$ and $y^{2}-6y=(y - 3)^{2}-9$.

Step3: Rewrite the original equation

The original equation $x^{2}+4x + y^{2}-6y=-4$ becomes $(x + 2)^{2}-4+(y - 3)^{2}-9=-4$.

Step4: Simplify the equation

$(x + 2)^{2}+(y - 3)^{2}-13=-4$, then $(x + 2)^{2}+(y - 3)^{2}=9$.

Step5: Identify the center and radius

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

Answer:

The center is located at $(-2,3)$, and the radius is $3$.