The point diametrically opposite to the point $P(1,0)$ on the circle $x^2+y^2+2 x+4 y-3=0$ is
The point diametrically opposite to the point $P(1,0)$ on the circle $x^2+y^2+2 x+4 y-3=0$ is
-
$(3,-4)$
-
$(-3,4)$
-
$(-3,-4)$
-
$(3,4)$
Solution
Centre $(-1,-2)$
Let $(\alpha, \beta)$ is the required point $\frac{\alpha+1}{2}=-1$ and $\frac{\beta+0}{2}=-2$.
Asked in: JEE Main 2008
Practice more Circle questions on Aicharya