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 B (-3,-4)

L$A$ be the required point with coordinate $(x,y)$.

Given equation of circle:-

$x^2+y^2+2x+4y-3=0$

To find:- Coordinate of point $A$

Now,

$x^2+y^2+2x+4y-3=0$

${(x+1)}^2+{(y+2)}^2-2-4-3=0$

${(x+1)}^2+{(y+2)}^2={\left(3\right)}^2$

From the above equation, the centre of the circle is $(-1,-2)$.

Since $AP$ is the diameter of the circle, the centre will be the mid-point of $AB$.

now, as centre is the mid-point of $AB$.

$x$-coordinate of centre $=\frac{x+1}{2}$

$y$-coordinate of centre $=\frac{y+0}{2}=\frac{y}{2}$

But the centre of circle is $(-1,-2)$.

Therefore,

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

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

Thus the coordinate of $A$ is $(-3,-4)$.

Hence the correct answer is $\left(B\right)(-3,-4)$.