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

Options

A. `(3,4)`
B. `(3,-4)`
C. `(-3,4)`
D. `(-3,-4)`

Answer

D

Explanation

Given equation can be rewritten as
`(x+1)^(2)+(y+2)^(2)=(2sqrt(2))^(2)`
Let required point be `Q(alpha, beta).` Then, mid-point of `P(1,0) and Q(alpha, beta)` is the centre of the circle.
i.e. `(alpha+1)/(2)= -1 and (beta+0)/(2)= -2`
`rArr alpha = -3 and beta = -4`
Hence, required point is `(-3,-4)`.