Circle And System Of Circles Question 209
For what value of k, the points (0, 0), (1, 3), (2, 4) and (k, 3) are concyclic
[RPET 1997]
Options:
2
1
4
5
Show Answer
Answer:
Correct Answer: B
Solution:
The equation of circle through points (0, 0), (1, 3) and (2, 4) is $ x^{2}+y^{2}-6x-4y=0 $
Point $ (k,,3) $ will be on the circle, if $ k^{2}+9-10k=0\Rightarrow k^{2}-10k+9=0 $
$ k^{2}-9k-k+9=0\Rightarrow k(k-9)-1(k-9)=0 $
$ \Rightarrow $ $ k=1 $ or $ k=9 $ .
BETA
