Circle And System Of Circles Question 314
Question: $ 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]
Options:
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 $
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Answer:
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( \frac{2a}{1+m^{2}},\frac{2am}{1+m^{2}} ) $ .
Hence the equation of circle (as chord OB as diameter) is $ (x^{2}+y^{2})(1+m^{2})-2a(x+my)=0 $ .
BETA
