Answer :
To solve the given problem and identify the center and radius of the circle from the equation [tex]\((x-6)^2 + (y-1)^2 = 9\)[/tex], follow these steps:
1. Identify the general form of the equation of a circle:
The general form of a circle's equation is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
2. Compare the given equation to the general form:
The given equation is [tex]\((x-6)^2 + (y-1)^2 = 9\)[/tex]. By comparing this with the general form [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], we can see that:
- [tex]\(h = 6\)[/tex]
- [tex]\(k = 1\)[/tex]
- [tex]\(r^2 = 9\)[/tex]
3. Identify the center of the circle:
The center [tex]\((h, k)\)[/tex] is [tex]\((6, 1)\)[/tex], as identified above.
4. Determine the radius of the circle:
The radius is found by taking the square root of [tex]\(r^2\)[/tex]. Since [tex]\(r^2 = 9\)[/tex], the radius [tex]\(r\)[/tex] is [tex]\(\sqrt{9} = 3.0\)[/tex].
So, the center of the circle is [tex]\((6, 1)\)[/tex] and the radius is [tex]\(3.0\)[/tex].
Now, fill in the table with the identified values:
[tex]\[ \begin{tabular}{|l|} \hline Center: (6, 1) \\ \hline Radius: 3.0 \\ \hline \\ \hline \end{tabular} \][/tex]
For graphing the circle:
- Plot the center of the circle at the point [tex]\((6, 1)\)[/tex].
- Since the radius is [tex]\(3.0\)[/tex], draw a circle with a fixed distance of [tex]\(3.0\)[/tex] units from the center in all directions.
Graphically, it should look like a circle centered at [tex]\((6, 1)\)[/tex] with all points on the circumference being [tex]\(3.0\)[/tex] units away from the center.
1. Identify the general form of the equation of a circle:
The general form of a circle's equation is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
2. Compare the given equation to the general form:
The given equation is [tex]\((x-6)^2 + (y-1)^2 = 9\)[/tex]. By comparing this with the general form [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], we can see that:
- [tex]\(h = 6\)[/tex]
- [tex]\(k = 1\)[/tex]
- [tex]\(r^2 = 9\)[/tex]
3. Identify the center of the circle:
The center [tex]\((h, k)\)[/tex] is [tex]\((6, 1)\)[/tex], as identified above.
4. Determine the radius of the circle:
The radius is found by taking the square root of [tex]\(r^2\)[/tex]. Since [tex]\(r^2 = 9\)[/tex], the radius [tex]\(r\)[/tex] is [tex]\(\sqrt{9} = 3.0\)[/tex].
So, the center of the circle is [tex]\((6, 1)\)[/tex] and the radius is [tex]\(3.0\)[/tex].
Now, fill in the table with the identified values:
[tex]\[ \begin{tabular}{|l|} \hline Center: (6, 1) \\ \hline Radius: 3.0 \\ \hline \\ \hline \end{tabular} \][/tex]
For graphing the circle:
- Plot the center of the circle at the point [tex]\((6, 1)\)[/tex].
- Since the radius is [tex]\(3.0\)[/tex], draw a circle with a fixed distance of [tex]\(3.0\)[/tex] units from the center in all directions.
Graphically, it should look like a circle centered at [tex]\((6, 1)\)[/tex] with all points on the circumference being [tex]\(3.0\)[/tex] units away from the center.
