To determine the center and radius of the circle given by the equation:
[tex]\[ x^2 + y^2 + 6x + 10y + 18 = 0 \][/tex]
We first need to rewrite this equation in the standard form of a circle's equation:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
To do this, we'll complete the square for both the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
1. Start with the given equation:
[tex]\[ x^2 + y^2 + 6x + 10y + 18 = 0 \][/tex]
2. Separate the equation into [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms:
[tex]\[ x^2 + 6x + y^2 + 10y + 18 = 0 \][/tex]
3. Complete the square for the [tex]\(x\)[/tex]-terms:
[tex]\[ x^2 + 6x \][/tex]
Add and subtract [tex]\((\frac{6}{2})^2 = 9\)[/tex]:
[tex]\[ (x + 3)^2 - 9 \][/tex]
4. Complete the square for the [tex]\(y\)[/tex]-terms:
[tex]\[ y^2 + 10y \][/tex]
Add and subtract [tex]\((\frac{10}{2})^2 = 25\)[/tex]:
[tex]\[ (y + 5)^2 - 25 \][/tex]
5. Rewrite the equation substituting the completed squares:
[tex]\[ (x + 3)^2 - 9 + (y + 5)^2 - 25 + 18 = 0 \][/tex]
6. Combine constants:
[tex]\[ (x + 3)^2 + (y + 5)^2 - 16 = 0 \][/tex]
[tex]\[ (x + 3)^2 + (y + 5)^2 = 16 \][/tex]
Now, we have the equation in the standard form:
[tex]\[ (x - (-3))^2 + (y - (-5))^2 = 4^2 \][/tex]
From this, we can see that:
- The center [tex]\((h, k)\)[/tex] of the circle is [tex]\((-3, -5)\)[/tex]
- The radius [tex]\(r\)[/tex] of the circle is [tex]\(4\)[/tex] units
So, the correct answers are:
The center of the circle is at [tex]\( (-3, -5) \)[/tex] and the radius of the circle is [tex]\( 4 \)[/tex] units.
