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

The end points of a diameter are always symmetric with respect to the centre of the circle. Hence, if we know the centre $$C(h,k)$$ and one end point $$P(x_1,y_1)$$, the diametrically opposite point $$Q(x_2,y_2)$$ is obtained from

$$2h = x_1 + x_2,\qquad 2k = y_1 + y_2$$

Step 1: Find the centre of the given circle.
The equation is $$x^2 + y^2 + 2x + 4y - 3 = 0$$.
Complete the squares:

$$x^2 + 2x + y^2 + 4y = 3$$
$$\bigl(x^2 + 2x + 1\bigr) + \bigl(y^2 + 4y + 4\bigr) = 3 + 1 + 4$$
$$(x+1)^2 + (y+2)^2 = 8$$

Thus the centre is $$C(-1,-2)$$.

Step 2: Use the midpoint (centre) formula.
Given point $$P(1,0)$$ and centre $$C(-1,-2)$$, let $$Q(x_2,y_2)$$ be the required opposite point.

For the $$x$$-coordinate:
$$2(-1) = 1 + x_2 \;\Longrightarrow\; x_2 = -3$$

For the $$y$$-coordinate:
$$2(-2) = 0 + y_2 \;\Longrightarrow\; y_2 = -4$$

Hence $$Q(-3,-4)$$ is diametrically opposite to $$P(1,0)$$.

Option C which is: ($$-3$$, $$-4$$)