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Q.
The point diametrically opposite to the point $P(1,0)$ on the circle $x^{2}+y^{2}+2 x+4 y-3=0$ is
Conic Sections
Solution:
The given circle is $x^{2}+y^{2}+2 x+4 y-3=0$
Centre $(-1,-2)$
Let $Q(\alpha, \beta)$ be the point diametrically opposite to
the point $P(1,0)$
then $\frac{1+\alpha}{2}=-1$ and $\frac{0+\beta}{2}=-2 $
$ \Rightarrow \alpha=-3, \beta=-4$
So, $Q$ is $(-3,-4)$
