To find the point that is diametrically opposite to the point [tex]\( P(1,0) \)[/tex] on the circle given by the equation [tex]\( x^2 + y^2 + 2x + 4y - 3 = 0 \)[/tex], we need to follow these steps:
1. Rewrite the circle equation in standard form:
The general 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 and [tex]\(r\)[/tex] is the radius. We need to rewrite the given equation in this form by completing the square.
2. Group and complete the square for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
The given equation is [tex]\( x^2 + y^2 + 2x + 4y - 3 = 0 \)[/tex].
- First, rearrange the equation: [tex]\(x^2 + 2x + y^2 + 4y = 3\)[/tex].
- Now, complete the square for the [tex]\(x\)[/tex] terms and the [tex]\(y\)[/tex] terms separately.
Completing the square for the [tex]\(x\)[/tex] terms:
[tex]\[
x^2 + 2x \quad \text{becomes} \quad (x + 1)^2 - 1
\][/tex]
Completing the square for the [tex]\(y\)[/tex] terms:
[tex]\[
y^2 + 4y \quad \text{becomes} \quad (y + 2)^2 - 4
\][/tex]
Substituting these back into the equation:
[tex]\[
(x + 1)^2 - 1 + (y + 2)^2 - 4 = 3
\][/tex]
Simplifying this, we get:
[tex]\[
(x + 1)^2 + (y + 2)^2 - 5 = 3
\][/tex]
Adding 5 to both sides, we get:
[tex]\[
(x + 1)^2 + (y + 2)^2 = 8
\][/tex]
Now the circle is in standard form with center [tex]\((-1, -2)\)[/tex] and radius [tex]\(\sqrt{8}\)[/tex].
3. Find the point diametrically opposite to [tex]\( P(1,0) \)[/tex]:
The coordinates of the point diametrically opposite to [tex]\( P(x_1, y_1) \)[/tex] on a circle with center [tex]\((h, k)\)[/tex] can be found using:
[tex]\[
P' = (2h - x_1, 2k - y_1)
\][/tex]
Here, [tex]\( (h,k) = (-1, -2) \)[/tex] and [tex]\( P(1,0) \)[/tex].
4. Calculate the coordinates of the opposite point:
[tex]\[
P' = (2(-1) - 1, 2(-2) - 0) = (-2 - 1, -4 - 0) = (-3, -4)
\][/tex]
So, the point diametrically opposite to [tex]\( P(1,0) \)[/tex] is [tex]\( \boxed{(-3, -4)} \)[/tex].
