A) (-1,-2)
B) (1,2)
C) (1,-2)
D) (-1,2)
Correct Answer: A
Solution :
: Since,\[{{x}_{1}},{{x}_{2}}\]are roots of \[{{x}^{2}}+2x-3=0\] \[\therefore \]\[{{x}_{1}}+{{x}_{2}}=-2\Rightarrow \frac{{{x}_{1}}+{{x}_{2}}}{2}=1\] Also, \[{{y}_{1}},{{y}_{2}}\]are roots of \[{{y}^{2}}+4y-12=0\] \[\therefore \]\[{{y}_{1}}+{{y}_{2}}=-4\Rightarrow \frac{{{y}_{1}}+{{y}_{2}}}{2}=-2\] Centre of circle\[\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)=(-1,-2)\]You need to login to perform this action.
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