# General Equation of the Circle

### General Equation of the Circle:

#### 1. General Form of the Circle Equation:

• The general equation of a circle is $A{x}^{2}+A{y}^{2}+Dx+Ey+F=0$.
• Conditions: For this equation to represent a circle, $A=B$ (coefficients of ${x}^{2}$ and ${y}^{2}$ terms should be equal).

#### 2. Center and Radius from the General Equation:

• Center: The center of the circle is given by $\left(-\frac{D}{2A},-\frac{E}{2A}\right)$.
• Radius: $r=\sqrt{\frac{{D}^{2}+{E}^{2}-4AF}{4{A}^{2}}}$.

#### 3. Example of General Equation of Circle:

Given the equation $3{x}^{2}+3{y}^{2}-4x+6y-5=0$, determine its center and radius.

Solution:

Comparing the equation $A{x}^{2}+A{y}^{2}+Dx+Ey+F=0$ with $3{x}^{2}+3{y}^{2}-4x+6y-5=0$:

• $A=3$, $D=-4$, $E=6$, and $F=-5$.

Using the formulas:

• Center: $\left(-\frac{D}{2A},-\frac{E}{2A}\right)=\left(-\frac{-4}{2\cdot 3},-\frac{6}{2\cdot 3}\right)=\left(\frac{2}{3},-1\right)$
• Radius: $r=\sqrt{\frac{\left(-4{\right)}^{2}+\left(6{\right)}^{2}-4\cdot 3\cdot \left(-5\right)}{4\cdot {3}^{2}}}=\sqrt{\frac{16+36+60}{36}}=\sqrt{\frac{112}{36}}=\sqrt{3.11}\approx 1.76$Therefore, the center of the circle represented by the equation $3{x}^{2}+3{y}^{2}-4x+6y-5=0$ is $\left(\frac{2}{3},-1\right)$, and its radius is approximately $1.76$.