Answer :
Sure, let's solve this step-by-step.
We start with the equation of the circle given in the standard form:
[tex]\[ (x - 5)^2 + y^2 = 81 \][/tex]
The standard form of the equation of a circle 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.
By comparing the given equation [tex]\((x - 5)^2 + y^2 = 81\)[/tex] with the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we can identify the following:
1. The term [tex]\((x - 5)^2\)[/tex] tells us that [tex]\(h = 5\)[/tex].
2. The term [tex]\(y^2\)[/tex] can be rewritten as [tex]\((y - 0)^2\)[/tex], which tells us that [tex]\(k = 0\)[/tex].
3. The right-hand side of the equation is [tex]\(81\)[/tex], which equals [tex]\(r^2\)[/tex]. Therefore, [tex]\(r^2 = 81\)[/tex].
To find the radius [tex]\(r\)[/tex], we take the square root of both sides:
[tex]\[ r = \sqrt{81} \][/tex]
Solving this, we get:
[tex]\[ r = 9 \][/tex]
Therefore, the radius of the circle is 9 units, and the center of the circle is at [tex]\((5, 0)\)[/tex].
Putting this information into the given format:
The radius of the circle is [tex]\(9\)[/tex] units.
The center of the circle is at [tex]\((5, 0)\)[/tex].
We start with the equation of the circle given in the standard form:
[tex]\[ (x - 5)^2 + y^2 = 81 \][/tex]
The standard form of the equation of a circle 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.
By comparing the given equation [tex]\((x - 5)^2 + y^2 = 81\)[/tex] with the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we can identify the following:
1. The term [tex]\((x - 5)^2\)[/tex] tells us that [tex]\(h = 5\)[/tex].
2. The term [tex]\(y^2\)[/tex] can be rewritten as [tex]\((y - 0)^2\)[/tex], which tells us that [tex]\(k = 0\)[/tex].
3. The right-hand side of the equation is [tex]\(81\)[/tex], which equals [tex]\(r^2\)[/tex]. Therefore, [tex]\(r^2 = 81\)[/tex].
To find the radius [tex]\(r\)[/tex], we take the square root of both sides:
[tex]\[ r = \sqrt{81} \][/tex]
Solving this, we get:
[tex]\[ r = 9 \][/tex]
Therefore, the radius of the circle is 9 units, and the center of the circle is at [tex]\((5, 0)\)[/tex].
Putting this information into the given format:
The radius of the circle is [tex]\(9\)[/tex] units.
The center of the circle is at [tex]\((5, 0)\)[/tex].