question_answer

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

A) \[\left( -3,-4 \right)\]

B)                        \[\left( 3,\,4 \right)\]     

C)        \[\left( 3,\,-4 \right)\]   

D)        \[\left( -3,\,\,4 \right)\]

Correct Answer: A

Solution :

                Given \[{{x}^{2}}+{{y}^{2}}+2x+4y-3=0\]; Centre \[\left( -1,-2 \right)\] C is the mid-point of P and Q so \[{{x}_{1}}=-3\] and \[{{y}_{1}}=-4\]                 \[\Rightarrow Q\left( -3,\,-4 \right)\]