Question

the equation of a circle is given below. identify the radius and the center. then graph the circle.
$x^{2}+y^{2}+4x - 6y = 3$

Explanation:

Step1: Complete the square for x and y terms

We have the equation $x^{2}+y^{2}+4x - 6y=3$.
For the x - terms: $x^{2}+4x=(x + 2)^{2}-4$.
For the y - terms: $y^{2}-6y=(y - 3)^{2}-9$.
So the equation becomes $(x + 2)^{2}-4+(y - 3)^{2}-9 = 3$.

Step2: Rewrite the equation in standard form

Simplify the above equation:
$(x + 2)^{2}+(y - 3)^{2}=3 + 4+9$.
$(x + 2)^{2}+(y - 3)^{2}=16$.
The standard form of a circle's equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius.

Step3: Identify the radius and center

Comparing $(x + 2)^{2}+(y - 3)^{2}=16$ with $(x - a)^{2}+(y - b)^{2}=r^{2}$, we have $a=-2$, $b = 3$ and $r^{2}=16$, so $r = 4$.

Answer:

Radius: 4
Center: $(-2,3)$