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

Correct option is C

Given:
Point P = (1, 0)
Circle: $$x^{2} + y^{2} + 2x + 4y − 3 = 0$$​​
Formula used:
For a circle, the midpoint of the endpoints of a diameter is the center.
If P(x₁, y₁) and Q(x₂, y₂) are diametrically opposite points, then
Center O = $$\left( \frac{x₁ + x₂}{2}, \frac{y₁ + y₂}{2} \right)$$​​
Solution:
Rewrite the circle in standard form:
​$$(x + 1)^{2} + (y + 2)^{2} = 8$$​​
So the center O = (−1, −2)
Let the required point be Q(x, y).
Using midpoint formula with P(1, 0):
​$$\frac{1 + x}{2} = −1\\\frac{0 + y}{2} = −2$$​​
Solving:
​$$1 + x = −2 => x = −3\\y = −4$$​​
So the diametrically opposite point is (−3, −4).
The correct answer is (c) $$(−3, −4).$$​​