A) \[(1+{{m}^{2}})({{x}^{2}}+{{y}^{2}})-2ax=0\]
B) \[(1+{{m}^{2}})({{x}^{2}}+{{y}^{2}})-2a(x+my)=0\]
C) \[(1+{{m}^{2}})({{x}^{2}}+{{y}^{2}})+2a(x+my)=0\]
D) \[(1+{{m}^{2}})({{x}^{2}}+{{y}^{2}})-2a(x-my)=0\]
Correct Answer: B
Solution :
Here the equation of circle is \[{{(x-a)}^{2}}+{{(y-0)}^{2}}={{a}^{2}}\Rightarrow {{x}^{2}}+{{y}^{2}}-2ax=0\] Now the point of intersection of circle and chord i.e., O and B are O(0, 0) and \[B\left( \frac{2a}{1+{{m}^{2}}},\frac{2am}{1+{{m}^{2}}} \right)\]. Hence the equation of circle (as chord OB as diameter) is \[({{x}^{2}}+{{y}^{2}})(1+{{m}^{2}})-2a(x+my)=0\].You need to login to perform this action.
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