Question
The value of k for (2k, 3k), (0, 0), (1, 0) and (0, 1) to be on the circle is
Options :
2/13
5/13
1/13
2/13
Answer :
5/13
Solution :
C: (x – 1)x + y(y – 1) = 0
x2 + y2 – x – y = 0
Now (2k, 3k) lies on C.
4k2 + 9k2 – 2k – 3k = 0
13k2 – 5k = 0
k(13k – 5) = 0
⇒ 5/13
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