Given the equation of a circle [tex]\( x^2 + (y - 10)^2 = 16 \)[/tex], we need to find the radius and the center of the circle.
First, let's recall the standard form of a circle's equation:
[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]\( x^2 + (y - 10)^2 = 16 \)[/tex] with the standard form:
1. The term [tex]\((x - h)^2\)[/tex] in the standard form corresponds to [tex]\( x^2\)[/tex] in the given equation. This means [tex]\( h = 0 \)[/tex].
2. The term [tex]\((y - k)^2\)[/tex] in the standard form corresponds to [tex]\((y - 10)^2\)[/tex] in the given equation. This means [tex]\( k = 10 \)[/tex].
So, the center of the circle is [tex]\((h, k) = (0, 10)\)[/tex].
Next, we need to find the radius [tex]\(r\)[/tex]. The right-hand side of the given equation is [tex]\( 16 \)[/tex], which corresponds to [tex]\( r^2 \)[/tex] in the standard form. Therefore,
[tex]\[ r^2 = 16 \][/tex]
To find [tex]\(r\)[/tex], we take the square root of both sides:
[tex]\[ r = \sqrt{16} = 4 \][/tex]
Thus, the radius of the circle is [tex]\( 4 \)[/tex] units.
To summarize:
- The radius of the circle is [tex]\( \boxed{4} \)[/tex] units.
- The center of the circle is at [tex]\( \boxed{(0, 10)} \)[/tex].
