Conic Sections Question 221
Question: Four distinct points $ (2k,3k),(1,0),(0,1) $ and $ (0,0) $ lie on a circle for
Options:
A) Only one value of k
B) $ 0<k<1 $
C) $ k<0 $
D) All integral values of k
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The equation of the circle through (1, 0), (0, 1) and (0, 0) is $ x^{2}+y^{2}-x-y=0 $ it passes through $ (2k,3k) $ So, $ 4k^{2}+9k^{2}-2k-3k=0 $ or $ 13k^{2}-5k=0 $
$ \Rightarrow k(13k-5)=0\Rightarrow k=0 $ or $ k=\frac{5}{13} $ But $ k\ne 0 $ [ $ \because $ all the four points are distinct]
$ \therefore k=\frac{5}{13}. $
BETA
