The point diametrically opposite to the point P(1, 0) on the circle `x^(2)+y^(2) + 2x+4y- 3 = 0` is

Options

A. (3 ,-4 )
B. (-3,4)
C. (-3 ,-4 )
D. (3,4).

Answer

To find the point diametrically opposite to the point \( P(1, 0) \) on the circle given by the equation \( x^2 + y^2 + 2x + 4y - 3 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation in Standard Form
First, we need to rewrite the given circle equation in standard form. The general equation of a circle is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \( (h, k) \) is the center and \( r \) is the radius.

The given equation is:

\[ x^2 + y^2 + 2x + 4y - 3 = 0 \]

We can rearrange it:

\[ x^2 + 2x + y^2 + 4y = 3 \]

Now, we will complete the square for both \( x \) and \( y \).

### Step 2: Complete the Square
For \( x^2 + 2x \):

\[ x^2 + 2x = (x + 1)^2 - 1 \]

For \( y^2 + 4y \):

\[ y^2 + 4y = (y + 2)^2 - 4 \]

Substituting these back into the equation gives:

\[ (x + 1)^2 - 1 + (y + 2)^2 - 4 = 3 \]

Simplifying this:

\[ (x + 1)^2 + (y + 2)^2 - 5 = 3 \]

\[ (x + 1)^2 + (y + 2)^2 = 8 \]

### Step 3: Identify the Center and Radius
From the standard form \( (x + 1)^2 + (y + 2)^2 = 8 \), we can identify the center \( (h, k) \) and the radius \( r \):

- Center \( (h, k) = (-1, -2) \)
- Radius \( r = \sqrt{8} = 2\sqrt{2} \)

### Step 4: Find the Diametrically Opposite Point
The point \( P(1, 0) \) is given. The diametrically opposite point \( P' \) can be found using the center of the circle. The formula to find the diametrically opposite point is:

\[ P' = (2h - x, 2k - y) \]

Substituting \( (h, k) = (-1, -2) \) and \( (x, y) = (1, 0) \):

\[ P' = (2(-1) - 1, 2(-2) - 0) \]

Calculating this gives:

\[ P' = (-2 - 1, -4 - 0) = (-3, -4) \]

### Final Answer
The point diametrically opposite to the point \( P(1, 0) \) on the circle is:

\[ \boxed{(-3, -4)} \]