question_answer

\[y=mx\] is a chord of a circle of radius a and the diameter of the circle lies along x-axis and one end of this chord in origin .The equation of the circle described on this chord as diameter is                                                                      [MP PET 1990]

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 \[({{x}^{2}}+{{y}^{2}})(1+{{m}^{2}})-2a(x+my)=0\].