Circle And System Of Circles Question 68
Question: From the origin chords are drawn to the circle $ {{(x-1)}^{2}}+y^{2}=1 $ . The equation of the locus of the middle points of these chords is
[IIT 1985; EAMCET 1991]
Options:
A) $ x^{2}+y^{2}-3x=0 $
B) $ x^{2}+y^{2}-3y=0 $
C) $ x^{2}+y^{2}-x=0 $
D) $ x^{2}+y^{2}-y=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
The given circle is $ x^{2}+y^{2}-2x=0 $ . Let $ (x_1,\ y_1) $ be the middle point of any chord of this circle, than its equation is $ S_1=T $ . or $ x_1^{2}+y_1^{2}-2x_1=xx_1+yy_1-(x+x_1) $
If it passes through (0, 0), then $ x_1^{2}+y_1^{2}-2x_1=-x_1\Rightarrow x_1^{2}+y_1^{2}-x_1=0 $
Hence the required locus of the given point $ (x_1,\ y_1) $ is $ x^{2}+y^{2}-x=0 $ .
BETA
