Question

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

• A. $(-3,-4)$
• B. $(3,4)$
• C. $(3,-4)$
• D. $(-3,4)$

Answer 1

The correct option is A $(-3,-4)$

Completing the squares of the equation $x^2+y^2+2x+4y-3=0$ as follows:
$(x^2+2x+1)-1+(y^2+4y+4)-4-3=0$
$(x+1{)}^2+(y+2{)}^2=8$

It represents a circle with center $(-1,-2)$ and radius $2\sqrt{2}$
Note that, $P(1,0)$ is on the circle, therefore,

$(1+1{)}^2+(0+2{)}^2=8$
Now, we need to find the diametrically opposite point $Q$ to point $P$ on this circle,

Here centre $O(-1,-2)$ will be the mid point of line joining $P$ and $Q$.

Lets, assume coordinates of be $Q\equiv (x,y)$

$-1=\frac{1+x}{2}\Rightarrow x=-3$

$-2=\frac{0+y}{2}\Rightarrow y=-4$

$\therefore$ Coordinates of $Q$ are $(-3,-4)$.