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

Options

A. (3, - 4)
B. (- 3, 4)
C. (1, 0)
D. (3, 4)

Answer

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

### Step 1: Rewrite the equation of the circle in standard form
The given equation is:
\[ x^2 + y^2 + 2x + 4y - 3 = 0 \]
We can rearrange it to complete the square for both \(x\) and \(y\).

1. Group the \(x\) terms and the \(y\) terms:
\[ (x^2 + 2x) + (y^2 + 4y) = 3 \]

2. Complete the square for \(x\):
\[ x^2 + 2x = (x + 1)^2 - 1 \]

3. Complete the square for \(y\):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

4. Substitute back into the equation:
\[ (x + 1)^2 - 1 + (y + 2)^2 - 4 = 3 \]
Simplifying gives:
\[ (x + 1)^2 + (y + 2)^2 - 5 = 3 \]
\[ (x + 1)^2 + (y + 2)^2 = 8 \]

Now, we have the standard form of the circle:
\[ (x + 1)^2 + (y + 2)^2 = 8 \]
This indicates that the center of the circle \(C\) is at \((-1, -2)\) and the radius is \(\sqrt{8}\).

### Step 2: Find the diametrically opposite point
Let the point \(P\) be \((-3, -4)\) and let the diametrically opposite point be \(Q(\alpha, \beta)\). The center \(C\) is the midpoint of \(P\) and \(Q\).

Using the midpoint formula:
\[ C = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the known values:
\[ (-1, -2) = \left( \frac{-3 + \alpha}{2}, \frac{-4 + \beta}{2} \right) \]

### Step 3: Set up equations
From the midpoint coordinates, we can set up two equations:
1. For the x-coordinates:
\[ \frac{-3 + \alpha}{2} = -1 \]
Multiplying both sides by 2:
\[ -3 + \alpha = -2 \implies \alpha = 1 \]

2. For the y-coordinates:
\[ \frac{-4 + \beta}{2} = -2 \]
Multiplying both sides by 2:
\[ -4 + \beta = -4 \implies \beta = 0 \]

### Step 4: Conclusion
The point diametrically opposite to \((-3, -4)\) is:
\[ Q(1, 0) \]

### Final Answer:
The correct option is (iii) \( (1, 0) \).