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 will follow these steps:

### Step 1: Convert the circle equation to standard form
The standard form 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.

Starting with the given equation:
\[ x^2 + y^2 + 2x + 4y - 3 = 0 \]
we can rearrange it as:
\[ x^2 + 2x + y^2 + 4y = 3 \]

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

### Step 2: Complete the square for \(x\)
For \(x^2 + 2x\), we take half of the coefficient of \(x\) (which is 2), square it, and add/subtract it:
\[ x^2 + 2x = (x + 1)^2 - 1 \]

### Step 3: Complete the square for \(y\)
For \(y^2 + 4y\), we take half of the coefficient of \(y\) (which is 4), square it, and add/subtract it:
\[ y^2 + 4y = (y + 2)^2 - 4 \]

### Step 4: Substitute back into the equation
Now substituting back into the equation:
\[ (x + 1)^2 - 1 + (y + 2)^2 - 4 = 3 \]
This simplifies to:
\[ (x + 1)^2 + (y + 2)^2 - 5 = 3 \]
or
\[ (x + 1)^2 + (y + 2)^2 = 8 \]

### Step 5: Identify the center and radius
From the standard form, we can identify:
- Center \((h, k) = (-1, -2)\)
- Radius \(r = \sqrt{8} = 2\sqrt{2}\)

### Step 6: Find the diametrically opposite point
The point \(P(1, 0)\) and the center \(O(-1, -2)\) will help us find the opposite point \(Q(x, y)\).

Using the midpoint formula, since \(O\) is the midpoint of \(P\) and \(Q\):
\[ O = \left(\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}\right) \]

Substituting the coordinates:
\[ (-1, -2) = \left(\frac{1 + x}{2}, \frac{0 + y}{2}\right) \]

### Step 7: Set up equations
From the x-coordinates:
\[ -1 = \frac{1 + x}{2} \implies -2 = 1 + x \implies x = -3 \]

From the y-coordinates:
\[ -2 = \frac{0 + y}{2} \implies -4 = y \]

### Step 8: Conclusion
Thus, the coordinates of the point \(Q\) diametrically opposite to \(P(1, 0)\) are:
\[ Q(-3, -4) \]

### Final Answer
The point diametrically opposite to \(P(1, 0)\) is \((-3, -4)\).

---