question_answer

                 If  \[({{x}_{1}},{{y}_{1}})\] & \[({{x}_{2}},{{y}_{2}})\] are the ends of diameter of a circle such that \[{{x}_{1}}\] & \[{{x}_{2}}\] are the roots of the equation \[a{{x}^{2}}+bx+c=0\] and  \[{{y}_{1}}\] & \[{{y}_{2}}\] are the roots of the equation; \[p{{y}^{2}}+qy+c=0\] Then the co-ordinates of the centre of the circle is:             

A) \[\left( \frac{b}{2a},\frac{q}{2p} \right)\]             

B) \[\left( -\frac{b}{2a},-\frac{q}{2p} \right)\]

C) \[\left( \frac{b}{a},\frac{q}{p} \right)\]                

D) None of these 

Correct Answer: B

Solution :

\[{{x}_{1}}+{{x}_{2}}=-b/q\] and \[{{y}_{1}}+{{y}_{2}}=-q/p\]

Center \[\left[ \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right]=\left[ -\frac{b}{2a},\frac{-\,q}{2p} \right]\]

[Hint: required equation is

\[{{x}^{2}}+{{y}^{2}}+\frac{b}{a}x+\frac{q}{p}y+\frac{c}{a}+\frac{c}{p}=0\Rightarrow B\]