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Question

Four distinct points (2k, 3k), (1, 0) (0, 1) and (0, 0) lie on a circle for

A
∀ k ∈ I
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B
k < 0
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C
0 < k < 1
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D
For two values of k
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Solution

The correct option is D For two values of k
General equation of circle is,
x2+y2+2gx+2fy+c=0

It passes through (0,0), (1, 0) and (0, 1); c = 0
Now 2g + 1 = 0 g = 12 and 2f + 1 = 0 f = 12
Hence equation of circle is x2+y2xy=0
Point (2k, 3k) lies on the circle
4k2+9k25k=0

13k25k=0
k=0 or k=513.


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