Question

given a circle with the equation x² + y² - 4x + 6y - 3 = 0, what is the coordinate of its center and the length of its radius? enter your answers in the boxes. the coordinate of the circles center is ( , ). the circles radius is units.

Explanation:

Step1: Rewrite the circle equation in standard form

The general equation of a circle is $x^{2}+y^{2}+Dx + Ey+F = 0$, and its standard - form is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center of the circle and $r$ is the radius. Given $x^{2}+y^{2}-4x + 6y-3 = 0$. Complete the square for $x$ and $y$ terms. 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 $(x - 2)^{2}-4+(y + 3)^{2}-9-3 = 0$.

Step2: Simplify the equation to standard form

Rearrange the equation: $(x - 2)^{2}+(y + 3)^{2}=4 + 9+3$.

Step3: Identify the center and radius

The standard - form of the circle is $(x - 2)^{2}+(y+3)^{2}=16$. The center of the circle $(a,b)=(2,-3)$ and the radius $r=\sqrt{16}=4$.

Answer:

The coordinate of the circle's center is $(2,-3)$. The circle's radius is $4$ units.