To find the center and radius of the circle given by the equation [tex]\( x^2 + 2x + y^2 = 35 \)[/tex], we can follow these steps:
1. Rewrite the given equation in standard form:
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 the radius.
2. Complete the square for the [tex]\( x \)[/tex] terms:
We need to make [tex]\( x^2 + 2x \)[/tex] look like a perfect square trinomial. To do this, we add and subtract the same value inside the equation:
[tex]\[
x^2 + 2x = (x + 1)^2 - 1
\][/tex]
Adding and subtracting 1 allows us to write [tex]\( x^2 + 2x \)[/tex] as a perfect square.
3. Substitute back into the equation:
Substitute the completed square back into the original equation:
[tex]\[
(x + 1)^2 - 1 + y^2 = 35
\][/tex]
4. Simplify the equation:
Move the constant term outside the perfect square to the right-hand side of the equation:
[tex]\[
(x + 1)^2 + y^2 = 36
\][/tex]
5. Identify the center and radius:
The equation is now in the standard form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\( r \)[/tex] is the radius. Comparing this with our equation:
[tex]\[
(x + 1)^2 + y^2 = 36
\][/tex]
We can see that:
- [tex]\( h = -1 \)[/tex]
- [tex]\( k = 0 \)[/tex]
- [tex]\( r \)[/tex] is the square root of 36, so [tex]\( r = 6 \)[/tex]
So, the center of the circle is [tex]\((-1, 0)\)[/tex] and the radius is [tex]\( 6 \)[/tex].
[tex]\[
\text{Center} \, (h, k) = (-1, 0)
\][/tex]
[tex]\[
\text{Radius} \, r = 6
\][/tex]
