Let's solve the problem step by step to determine the center and radius of the given circle:
The given equation of the circle is:
[tex]\[ x^2 + y^2 + 6x + 10y + 18 = 0 \][/tex]
1. Rewriting the equation in the standard form by completing the square:
First, we group the \( x \) terms together and the \( y \) terms together:
[tex]\[ (x^2 + 6x) + (y^2 + 10y) = -18 \][/tex]
2. Completing the square for the \( x \) terms:
We take the coefficient of \( x \), which is 6, halve it to get 3, and then square it to get 9. Add and subtract 9 inside the \( x \) terms:
[tex]\[ (x^2 + 6x + 9 - 9) + (y^2 + 10y) = -18 \][/tex]
This can be rewritten as:
[tex]\[ ((x + 3)^2 - 9) + (y^2 + 10y) = -18 \][/tex]
3. Completing the square for the \( y \) terms:
We take the coefficient of \( y \), which is 10, halve it to get 5, and then square it to get 25. Add and subtract 25 inside the \( y \) terms:
[tex]\[ ((x + 3)^2 - 9) + ( (y + 5)^2 - 25 ) = -18 \][/tex]
This can be rewritten as:
[tex]\[ (x + 3)^2 - 9 + (y + 5)^2 - 25 = -18 \][/tex]
4. Simplifying by combining constants:
Combine the constants on the right-hand side:
[tex]\[ (x + 3)^2 + (y + 5)^2 - 34 = -18 \][/tex]
Adding 34 to both sides results in:
[tex]\[ (x + 3)^2 + (y + 5)^2 = 16 \][/tex]
5. Standard form:
Now, the equation is in the standard form of a circle:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
From the equation \((x + 3)^2 + (y + 5)^2 = 16\):
- The center \((h, k)\) is \((-3, -5)\).
- The radius \(r\) is \(\sqrt{16} = 4\).
So, the center of the circle is at \((-3, -5)\) and the radius of the circle is 4 units.
To fill in the boxes:
The center of the circle is at [tex]\((-3, -5)\)[/tex] and the radius of the circle is 4 units.
