To solve the given equation [tex]\((x+1)^2 + (y-3)^2 = \sqrt{8}\)[/tex], let's recognize that this represents a geometric shape on the coordinate plane. Specifically, this form resembles the general equation of a circle, but not in its standard form due to the right-hand side being [tex]\(\sqrt{8}\)[/tex].
1. Standard Equation of a Circle:
The standard 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 its radius.
2. Transforming the Given Equation:
First, let's deal with the right-hand side of the given equation [tex]\( (x + 1)^2 + (y - 3)^2 = \sqrt{8} \)[/tex].
Since we know that for a circle, the equation includes the radius squared ([tex]\(r^2\)[/tex]), we should square [tex]\(\sqrt{8}\)[/tex] to match this form:
[tex]\[
(\sqrt{8})^2 = 8
\][/tex]
Thus, we rewrite the equation:
[tex]\[
(x + 1)^2 + (y - 3)^2 = 8
\][/tex]
3. Identify the Components:
Now the equation [tex]\((x + 1)^2 + (y - 3)^2 = 8\)[/tex] is in the standard circle form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. By comparing both forms, we can identify:
- The center of the circle [tex]\((h, k)\)[/tex] is [tex]\((-1, 3)\)[/tex], since [tex]\((x + 1)\)[/tex] indicates a center at [tex]\(x = -1\)[/tex] and [tex]\((y - 3)\)[/tex] indicates a center at [tex]\(y = 3\)[/tex].
- The radius [tex]\(r\)[/tex] can be found by taking the square root of 8:
[tex]\[
r = \sqrt{8} = 2\sqrt{2}
\][/tex]
4. Summary:
Therefore, the given equation represents a circle with:
- Center at [tex]\((-1, 3)\)[/tex]
- Radius [tex]\(2\sqrt{2}\)[/tex]
So, the final result is:
- The circle is centered at [tex]\((-1, 3)\)[/tex].
- The radius of the circle is [tex]\(2\sqrt{2}\)[/tex].
