If the point diametrically opposite to point $P (1, 0)$ on the circle $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$ is $(a, b)$ , then

| a | = 3

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| b | = 4

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| a | = 4

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| b | = 3

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Solution

The correct option is B $| b | = 4$
Given circle is $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$
Centre $= (- 1, - 2)$
Let $Q (a, b)$ be the point diametrically opposite to the point $P (1, 0)$ ,
$\frac{1 + a}{2} = - 1, \frac{0 + b}{2} = - 2 \Rightarrow a = - 3, b = - 4$

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