The point diametrically opposite to the point | vedclass

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(B) The given equation of the circle is $x^2 + y^2 + 2x + 4y - 3 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g

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$(-3, -4)$

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(B) The given equation of the circle is $x^2 + y^2 + 2x + 4y - 3 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = 2 \Rightarrow g = 1$ and $2f = 4 \Rightarrow f = 2$.The center of the circle is $(-g, -f) = (-1, -2)$.Let $Q(\alpha, \beta)$ be the point diametrically opposite to the point $P(1, 0)$.Since the center is the midpoint of the diameter $PQ$,we have:$\frac{1 + \alpha}{2} = -1$ $\Rightarrow 1 + \alpha = -2$ $\Rightarrow \alpha = -3$$\frac{0 + \beta}{2} = -2 \Rightarrow \beta = -4$Thus,the point $Q$ is $(-3, -4)$.

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

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