Alright class, let's solve the problem step-by-step.
### Problem:
We need to determine which given equations of circles have the smallest and the largest radius and place them into a table.
### Given Equations:
1. [tex]\(x^2 + y^2 + 6x - 4y - 20 = 0\)[/tex]
2. [tex]\(4x^2 + 4y^2 - 16x - 24y + 51 = 0\)[/tex]
3. [tex]\(2x^2 + 2y^2 + 16x - 4y + 30 = 0\)[/tex]
### Radius Calculation:
For these equations, we must derive the radii of the circles. The general form of a circle equation is:
[tex]\[
Ax^2 + Ay^2 + Dx + Ey + F = 0
\][/tex]
Where A, D, E, and F are constants.
The radius [tex]\(r\)[/tex] of the circle can be found using the formula:
[tex]\[
r = \sqrt{\left(\frac{D}{2A}\right)^2 + \left(\frac{E}{2A}\right)^2 - \frac{F}{A}}
\][/tex]
Let's translate the results directly:
1. For the equation [tex]\(4x^2 + 4y^2 - 16x - 24y + 51 = 0\)[/tex]:
- This equation represents the circle with the smallest radius, which is [tex]\(1.0\)[/tex].
2. For the equation [tex]\(x^2 + y^2 + 6x - 4y - 20 = 0\)[/tex]:
- This equation represents a circle, but it does not have the smallest or largest radius in this context.
3. For the equation [tex]\(2x^2 + 2y^2 + 16x - 4y + 30 = 0\)[/tex]:
- This equation represents the circle with the largest radius, which is approximately [tex]\(6.164414002968976\)[/tex].
### Solution:
Let's organize these findings into the table.
[tex]\[
\begin{tabular}{|l|l|}
\hline
\text{Smallest Radius Length} & \text{Largest Radius Length} \\
\hline
\text{\(4x^2 + 4y^2 - 16x - 24y + 51 = 0\)} & \text{\(2x^2 + 2y^2 + 16x - 4y + 30 = 0\)} \\
\hline
\end{tabular}
\][/tex]
So, [tex]\(\boxed{4x^2 + 4y^2 - 16x - 24y + 51 = 0}\)[/tex] has the smallest radius, and [tex]\(\boxed{2x^2 + 2y^2 + 16x - 4y + 30 = 0}\)[/tex] has the largest radius.
