Circle And System Of Circles Question 228
Question: Four distinct points $ (2k,,3k),(1,0)(0,1) $ and $ (0,0) $ lie on a circle for
[DCE 2005]
Options:
A) $ ,k\in I $
B) $ k<0 $
C) $ 0<k<1 $
D) For two values of k
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Answer:
Correct Answer: D
Solution:
General equation of circle is, $ x^{2}+y^{2}+2gx+2fy+c=0 $ It passes through (0,0), (1, 0) and (0, 1); \ $ c=0 $ Now $ 2g+1=0\Rightarrow g=-\frac{1}{2} $ and $ 2f+1=0\Rightarrow f=\frac{-1}{2} $
Hence equation of circle is $ x^{2}+y^{2}-x-y=0 $ Point $ (2k,3k) $ lies on the circle \ $ 4k^{2}+9k^{2}-5k=0 $
therefore $ 13k^{2}-5k=0 $
therefore $ k=0 $ or $ k=\frac{5}{13} $ .
BETA
