A) \[{{x}^{2}}+{{y}^{2}}+2ax+2cy+b+d=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2ax-2cy-b-d=0\]
C) \[{{x}^{2}}+{{y}^{2}}+2ax+2cy-b-d=0\]
D) \[{{x}^{2}}+{{y}^{2}}-2ax-2cy+b+d=0\]
Correct Answer: A
Solution :
Let the roots of equations \[{{x}^{2}}+2ax+b=0\] and \[{{x}^{2}}+2cx+d=0\] are\[({{\alpha }_{1}},\,\,{{\beta }_{1}})\]and\[({{\alpha }_{2}},\,\,{{\beta }_{2}})\] respectively. Then, \[{{\alpha }_{1}}+{{\beta }_{1}}=-2\alpha \] \[{{\alpha }_{1}}{{\beta }_{1}}=b\] \[{{\alpha }_{2}}+{{\beta }_{2}}=-2c\] \[{{\alpha }_{2}}{{\beta }_{2}}=d\] Thus, we have\[A({{\alpha }_{1}},\,\,{{\alpha }_{2}})\]and\[B({{\beta }_{1}},\,\,{{\beta }_{2}})\] Now, equations of circle with\[A({{\alpha }_{1}},\,\,{{\alpha }_{2}})\]and \[B({{\beta }_{1}},\,\,{{\beta }_{2}})\]as the diameter of the circle is \[(x-{{\alpha }_{1}})(x-{{\beta }_{1}})+(y-{{\alpha }_{2}})(y-{{\beta }_{2}})=0\] \[\Rightarrow \]\[{{x}^{2}}-({{\alpha }_{1}}+{{\beta }_{1}})x+{{\alpha }_{1}}{{\beta }_{1}}+{{y}^{2}}-({{\alpha }_{2}}+{{\beta }_{2}})y\] \[+{{\alpha }_{2}}{{\beta }_{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}+2ax+b+{{y}^{2}}+2cy+d=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+2ax+2cy+b+d=0\]which is the required equation of circle.
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