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)`

Explanation

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 will follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the given circle equation in standard form. The equation is:
\[ x^2 + y^2 + 2x + 4y - 3 = 0 \]
We can rearrange this to:
\[ x^2 + 2x + y^2 + 4y = 3 \]

### Step 2: Complete the Square
Next, we will complete the square for both \( x \) and \( y \).

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

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

Substituting these back into the equation gives:
\[ (x + 1)^2 - 1 + (y + 2)^2 - 4 = 3 \]
Simplifying this, we have:
\[ (x + 1)^2 + (y + 2)^2 - 5 = 3 \]
\[ (x + 1)^2 + (y + 2)^2 = 8 \]

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

### Step 4: Find the Point Diametrically Opposite to \( P(1, 0) \)
The point \( P(1, 0) \) lies on the circle. The center of the circle is \( C(-1, -2) \). The point diametrically opposite to \( P \) can be found using the midpoint formula. The midpoint \( M \) between \( P(1, 0) \) and the opposite point \( Q(a, b) \) is the center \( C(-1, -2) \).

Using the midpoint formula:
\[ M_x = \frac{x_1 + x_2}{2} \quad \text{and} \quad M_y = \frac{y_1 + y_2}{2} \]
We have:
\[ -1 = \frac{1 + a}{2} \quad \text{and} \quad -2 = \frac{0 + b}{2} \]

### Step 5: Solve for \( a \) and \( b \)
From the first equation:
\[ -1 = \frac{1 + a}{2} \implies -2 = 1 + a \implies a = -3 \]

From the second equation:
\[ -2 = \frac{0 + b}{2} \implies -4 = b \]

### Final Answer
Thus, the point diametrically opposite to \( P(1, 0) \) is:
\[ Q(-3, -4) \]