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

The given circle equation is $x^2+y^2+2x+4y-3=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$:$2g=2⟹g=1$$2f=4⟹f=2$ Centre $C=(-g,-f)=(-1,-2)$.Let the diametrically opposite point be $Q(\alpha ,\beta )$. The centre $C$ is the midpoint of $PQ$, where $P=(1,0)$.
$$\frac{\alpha +1}{2}=-1⟹\alpha +1=-2⟹\alpha =-3$$
$$\frac{\beta +0}{2}=-2⟹\beta =-4$$
So, the point is $(-3,-4)$.